Cherry has a binary matrix $A$ consisting of $N$ rows and $M$ columns. The rows are numbered from $1$ to $N$, columns are numbered from $1$ to $M$. Element at row $i$ ($1$ ≤ $i$ ≤ $N$) and column $j$ ($1$ ≤ $j$ ≤ $M$) is denoted as $A_{ij}$. All elements of $A$ are either $0$ or $1$.
He performs $Q$ queries on matrix. Each query is provided by four integers $x_{1}$, $y_{1}$, $x_{2}$, $y_{2}$ which define the rectangle, where ($x_{1}$, $y_{1}$) stands for the coordinates of the top left cell of the rectangle, while ($x_{2}$, $y_{2}$) stands for the coordinates of the bottom right cell. You need to flip all the bits i.e. ($0$ to $1$, $1$ to $0$) that are located fully inside the query rectangle.
Finally, print the matrix after performing all the queries.
Note: $x_{1}$ represents the row number while $y_{1}$ represents the column number.

-----Input:-----
- The first line of the input contains two integers $N$ and $M$ — the number of rows and the number of columns in the matrix.
- Each of the next $N$ lines contains a string of length $M$, where the $j^{th}$ character of $i^{th}$ line denotes the value of $A_{i,j}$. 
- Next line contains an integer $Q$ — the number of queries.
- Then follow $Q$ lines with queries descriptions. Each of them contains four space-seperated integers $x_{1}$, $y_{1}$, $x_{2}$, $y_{2}$ — coordinates of the up left and bottom right cells of the query rectangle.

-----Output:-----
Print the matrix, in the form of $N$ strings, after performing all the queries.

-----Constraints-----
- $1 \leq N,M \leq 1000$
- $0 \leq  A_{ij}  \leq 1$
- $1 \leq Q \leq 10^6$
- $1 \leq x_{1} \leq x_{2} \leq N$
- $1 \leq y_{1} \leq y_{2} \leq M$

-----Sample Input:-----
2 2
00
00
3
1 1 1 1
2 2 2 2
1 1 2 2 

-----Sample Output:-----
01
10

-----Explanation:-----
Example case 1:
After processing the 1st query 1 1 1 1, matrix becomes:
[1000][1000]\begin{bmatrix} 10 \\ 00  \end{bmatrix}
After processing the 2nd query 2 2 2 2, the matrix becomes:
[1001][1001]\begin{bmatrix} 10 \\ 01  \end{bmatrix}
After processing the 3rd query 1 1 2 2, matrix becomes:
[0110][0110]\begin{bmatrix} 01 \\ 10  \end{bmatrix}
We need to output the matrix after processing all queries.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'r,c=list(map(int,input().split()))\nA=[] #Given Matrix\ndp=[[0 for i in range(c+1)] for j in range(r+1)]\nfor i in range(r):\n A.append([int(j) for j in list(input())])\n #print(A)\n\nq=int(input())\n\n#print(A,dp)\nfor i in range(q):\n x1,y1,x2,y2=list(map(int,input().split()))\n #print(x1,y1,x2,y2)\n x1,y1,x2,y2=x1-1,y1-1,x2-1,y2-1\n dp[x1][y1]+=1\n dp[x2+1][y2+1]+=1\n dp[x1][y2+1]-=1\n dp[x2+1][y1]-=1\n\nfor i,v1 in enumerate(A):\n for j,v2 in enumerate(v1):\n  if i==0 and j==0:\n   pass\n  elif i==0:\n   dp[i][j]+=dp[i][j-1]\n  elif j==0:\n   dp[i][j]+=dp[i-1][j]\n  else:\n   dp[i][j]+=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]\n  #print(A,i,j)\n\n  if dp[i][j]%2!=0:\n   A[i][j]=1-A[i][j]\nfor i in A:\n for j in i:\n  print(j,end="")\n print()\n\n\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14021