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>
using namespace std;

void solve() {
	int n;
	cin >> n;

	string a[n], b[n];

	for(int i = 0; i < n; i++)
		cin >> a[i];

	for(int i = 0; i < n; i++)
		cin >> b[i];

	string c[n];
	for(int i = 0; i < n; i++)
		c[i] = a[i];

	for(int i = 0; i < n; i++) {
		if(a[0][i] != b[0][i]) {
			for(int j = 0; j < n; j++)
				c[j][i] = '1' - (c[j][i] - '0');
		}
	}

	bool ok = 1;

	for(int i = 0; i < n; i++) {
		bool ok2 = 0;
		if(c[i] == b[i])
			ok2 = 1;

		string tmp = c[i];
		for(int i = 0; i < n; i++)
			tmp[i] = '1' - (tmp[i] - '0');
		if(tmp == b[i])
			ok2 = 1;
		ok &= ok2;
	}

	if(ok) {
		cout << "YES" << '\n';
		return;
	}

	ok = 1;

	for(int i = 0; i < n; i++)
		c[i] = a[i];

	for(int i = 0; i < n; i++) {
		c[0][i] = '1' - (c[0][i] - '0');
		if(c[0][i] != b[0][i]) {
			for(int j = 0; j < n; j++)
				c[j][i] = '1' - (c[j][i] - '0');
		}
	}

	for(int i = 0; i < n; i++) {
		bool ok2 = 0;
		if(c[i] == b[i])
			ok2 = 1;
		string tmp = c[i];
		for(int i = 0; i < n; i++)
			tmp[i] = '1' - (tmp[i] - '0');
		if(tmp == b[i])
			ok2 = 1;

		ok &= ok2;
	}

	if(ok) {
		cout << "YES" << '\n';
		return;
	}

	cout << "NO" << '\n';

}

int main() {
	int t;
	cin >> t;
	while(t--)
		solve();	
}

/*

*/