inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1157/G:
You are given a binary matrix $a$ of size $n \times m$. A binary matrix is a matrix where each element is either $0$ or $1$.

You may perform some (possibly zero) operations with this matrix. During each operation you can inverse the row of this matrix or a column of this matrix. Formally, inverting a row is changing all values in this row to the opposite ($0$ to $1$, $1$ to $0$). Inverting a column is changing all values in this column to the opposite.

Your task is to sort the initial matrix by some sequence of such operations. The matrix is considered sorted if the array $[a_{1, 1}, a_{1, 2}, \dots, a_{1, m}, a_{2, 1}, a_{2, 2}, \dots, a_{2, m}, \dots, a_{n, m - 1}, a_{n, m}]$ is sorted in non-descending order.


-----Input-----

The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 200$) — the number of rows and the number of columns in the matrix.

The next $n$ lines contain $m$ integers each. The $j$-th element in the $i$-th line is $a_{i, j}$ ($0 \le a_{i, j} \le 1$) — the element of $a$ at position $(i, j)$.


-----Output-----

If it is impossible to obtain a sorted matrix, print "NO" in the first line.

Otherwise print "YES" in the first line. In the second line print a string $r$ of length $n$. The $i$-th character $r_i$ of this string should be '1' if the $i$-th row of the matrix is inverted and '0' otherwise. In the third line print a string $c$ of length $m$. The $j$-th character $c_j$ of this string should be '1' if the $j$-th column of the matrix is inverted and '0' otherwise. If there are multiple answers, you can print any.


-----Examples-----
Input
2 2
1 1
0 1

Output
YES
00
10

Input
3 4
0 0 0 1
0 0 0 0
1 1 1 1

Output
YES
010
0000

Input
3 3
0 0 0
1 0 1
1 1 0

Output
NO

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
input = sys.stdin.readline

n,m=list(map(int,input().split()))
A=[list(map(int,input().split())) for i in range(n)]

for i in range(m):
    #一行目をi-1まで0にする

    ANSR=[0]*n
    ANSC=[0]*m

    for j in range(i):
        if A[0][j]==1:
            ANSC[j]=1

    for j in range(i,m):
        if A[0][j]==0:
            ANSC[j]=1

    for r in range(1,n):
        B=set()
        for c in range(m):
            if ANSC[c]==0:
                B.add(A[r][c])
            else:
                B.add(1-A[r][c])

        if len(B)>=2:
            break
        if max(B)==0:
            ANSR[r]=1

    else:
        print("YES")
        print("".join(map(str,ANSR)))
        print("".join(map(str,ANSC)))
        return

ANSR=[0]*n
ANSC=[0]*m

for j in range(m):
    if A[0][j]==1:
        ANSC[j]=1

flag=0
for r in range(1,n):
    if flag==0:
        B=[]
        for c in range(m):
            if ANSC[c]==0:
                B.append(A[r][c])
            else:
                B.append(1-A[r][c])

        if max(B)==0:
            continue
        elif min(B)==1:
            ANSR[r]=1
            continue
        else:
            OI=B.index(1)
            if min(B[OI:])==1:
                flag=1
                continue

            OO=B.index(0)
            if max(B[OO:])==0:
                flag=1
                ANSR[r]=1
                continue

            else:
                print("NO")
                return

    else:

        B=set()
        for c in range(m):
            if ANSC[c]==0:
                B.add(A[r][c])
            else:
                B.add(1-A[r][c])

        if len(B)>=2:
            break
        if max(B)==0:
            ANSR[r]=1

else:
    print("YES")
    print("".join(map(str,ANSR)))
    print("".join(map(str,ANSC)))
    return

print("NO")