Unusual Matrix

You are given two binary square matrices a and b of size n×n. A matrix is called binary if each of its elements is equal to 0 or 1. You can do the following operations on the matrix a arbitrary number of times (0 or more):

• vertical xor. You choose the number j (1≤j≤n) and for all i (1≤i≤n) do the following: \(a_{i,j}:=ai,j⊕1\) (⊕ — is the operation xor (exclusive or)).
• horizontal xor. You choose the number i (1≤i≤n) and for all j (1≤j≤n) do the following: \(ai,j:=ai,j⊕1\).

Note that the elements of the a matrix change after each operation.

For example, if n=3 and the matrix a is:

\[\begin{pmatrix}1&1&0\\0&0&1\\1&1&0\end{pmatrix} \]

Then the following sequence of operations shows an example of transformations:

• vertical xor, j=1.

  \[a=\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix} \]

• horizontal xor,i=2.

  \[a=\begin{pmatrix}0&1&0\\0&1&0\\0&1&0\end{pmatrix} \]

• vertical xor, j=2.

  \[a=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix} \]

Check if there is a sequence of operations such that the matrix a becomes equal to the matrix b.

Input

The first line contains one integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤1000) — the size of the matrices.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix a.

An empty line follows.

The following n lines contain strings of length n, consisting of the characters '0' and '1' — the description of the matrix b.

It is guaranteed that the sum of n over all test cases does not exceed 1000.

Output

For each test case, output on a separate line:

• "YES", there is such a sequence of operations that the matrix a becomes equal to the matrix b;
• "NO" otherwise.

Example

input

3
3
110
001
110

000
000
000
3
101
010
101

010
101
010
2
01
11

10
10

output

YES
YES
NO

Note

The first test case is explained in the statements.

In the second test case, the following sequence of operations is suitable:

• horizontal xor, i=1;
• horizontal xor, i=2;
• horizontal xor, i=3;

It can be proved that there is no sequence of operations in the third test case so that the matrix a becomes equal to the matrix b.

设p[i]: 第i行异或次数, p[n+j]第j列异或次数

显然\(b[ i][j]=p[i]\oplus p[n+j]\oplus a[i][j]\)

有2n个变量，可以列出n*n个等式

可以发现，如果确定了\(p[n+n]\)，则可以确定其余所有的p[i]

不过只满足其中2n-1个等式

\(p[i]=a[i][n]\oplus b[i][n]\oplus p[n+n]\)

\(p[n+j]=a[n][n]\oplus b[n][n]\oplus a[n][j]\oplus b[n][j]\oplus p[n+n]\)

最后验证所有的等式，如果p[n+n]修改次数超过1次，则无解

int n,m,k;
char a[MAXN][MAXN],b[MAXN][MAXN];
int p[MAXN*2];
int main(){
    int t;read(t);
    while(t--){
        read(n);
        for(int i=1;i<=n;++i)read(a[i]+1);
        for(int i=1;i<=n;++i)read(b[i]+1);
        for(int i=1;i<=n;++i){
            p[i]=(a[i][n]-48)^(b[i][n]-48);
        }
        for(int j=1;j<=n;++j){
            p[n+j]=(a[n][j]-48)^(b[n][j]-48)^(a[n][n]-48)^(b[n][n]-48);
        }
        int f=0,chd=0;
        for(int i=1;i<=n;++i){
            for(int j=1;j<=n;++j){
                int k1=p[i]^p[n+j];
                int k2=(a[i][j]-48)^(b[i][j]-48);
                if(k1^f!=k2){
                    f^=1;
                    chd++;
                }
            }
        }
        if(chd<=1){
            puts("YES");
        }else{
            puts("NO");
        }
    }
    return 0;
}

标签：Unusual,xor,Matrix,following,pmatrix,test,oplus,matrix
来源： https://www.cnblogs.com/foursmonth/p/14332059.html