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.
// time-limit: 2000
#include<bits/stdc++.h>
using namespace std;

void solve() {
    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(n)), b(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < s.length(); j++) {
            a[i][j] = s[j] == '1';
        }
    }
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < s.length(); j++) {
            b[i][j] = s[j] == '1';
        }
    }
    vector<int> row(n), col(n);
    for (int val : {0, 1}) {
        row[0] = val;
        for (int j = 0; j < n; j++) {
            col[j] = (a[0][j] ^ b[0][j]) ^ row[0];
        }
        bool ok = true;
        for (int i = 1; i < n; i++) {
            row[i] = (a[i][0] ^ b[i][0] ^ col[0]);
            for (int j = 1; j < n; j++) {
                if ((a[i][j] ^ b[i][j]) != (row[i] ^ col[j])) {
                    ok = false;
                }
            }
        }
        if (ok) {
            cout << "YES\n";
            return;
        }
    }
    cout << "NO\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1, t = 0;
    cin >> T;
    while (t++ < T) {
        //cout << "Case #" << t << ": ";
        solve();
    }

    return 0;
}
