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<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<stdio.h>
#include<map>
#include<queue>
using namespace std;
#define Check(x) cout<<"x:"<<x<<" "
#define Min(x,y,z) min(x,min(z,y))
#define Max(x,y,z) max(x,max(z,y))
typedef long long ll;
const int MAXN = 1e3 + 5;
const int MAXM = 1e6 + 5;
const ll mod = 1e9 + 7;
ll n, m;
int mp[MAXN][MAXN];
int ver[MAXN];
int hor[MAXN];
int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        for (int i = 1;i <= n;i++) {
            for (int j = 1;j <= n;j++) {
                char ch;
                cin >> ch;
                mp[i][j] = ch - '0';
            }
        }
        for (int i = 1;i <= n;i++) {
            for (int j = 1;j <= n;j++) {
                char ch;
                cin >> ch;
                mp[i][j] = (mp[i][j] + (ch - '0')) % 2;
            }
        }
        if (n == 1) {
            cout << "YES" << endl;
            continue;
        }
        bool f = true;
        for (int i = 1;i <= n;i++) {
            if (mp[1][i] == 1) {
                hor[i] = 1;
            }
        }
        for (int i = 2;i <= n;i++) {
            if (hor[1])mp[i][1] = mp[i][1] ^ 1;
            if (mp[i][1] == 1)ver[i] = 1;
            else ver[i] = 0;
            for (int j = 2;j <= n;j++) {
                for (int k = 1;k <= ver[i] + hor[j];k++) {
                    mp[i][j] = mp[i][j] ^ 1;
                }
                if (mp[i][j] == 1) { f = false;break; }
            }
            if (!f)break;
        }
        for (int i = 1;i <= n;i++) {
            ver[i] = hor[i] = 0;
        }
        if (f)cout << "YES" << endl;
        else cout << "NO" << endl;
    }

}