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} := a_{i, j} ⊕ 1 (⊕ — is the operation [xor](https://en.wikipedia.org/wiki/Exclusive_or) (exclusive or)). 
  * horizontal xor. You choose the number i (1 ≤ i ≤ n) and for all j (1 ≤ j ≤ n) do the following: a_{i, j} := a_{i, 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. 



You can output "YES" and "NO" in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

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.
#include<bits/stdc++.h>
#define endl '\n'
#define bug puts("aaa");
#define Case int t;scanf("%d",&t);while(t--)
#define ios ios::sync_with_stdio(0),cin.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e3+10;

int n;
int a[N][N],b[N][N];

int check() {
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=n;j++) {
            if (a[i][j]^a[0][j]^a[i][0] != b[i][j]) return 0;
        }
    }
    return 1;
}

void run() {
    scanf("%d",&n);
    for (int i=0;i<=n;i++) {
        for (int j=0;j<=n;j++) {
            a[i][j] = 0;
        }
    }

    for (int i=1;i<=n;i++) {
        for (int j=1;j<=n;j++) {
            scanf("%1d",&a[i][j]);
        }
    }
    for (int i=1;i<=n;i++) {
        for (int j=1;j<=n;j++) {
            scanf("%1d",&b[i][j]);
        }
    }


    for (int k=2;k>0;k--) {
        if (k==1) {
            for (int i=1;i<=n;i++) {
                a[1][i] ^= 1;
            }
        }
        for (int i=1;i<=n;i++) {
            if (a[1][i]!=b[1][i]) {
                a[0][i] = 1;
            }
        }
        for (int i=k;i<=n;i++) {
            if (a[i][1]^a[0][1]!=b[i][1]) {
                a[i][0] = 1; 
            }
        }
        if (check()) {
            puts("YES");
            return;
        }
    }
    puts("NO");

}

int main() {
    
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    Case
    run();
    return 0;
}