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 <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cmath>
#include <algorithm>
#include <climits>
#include <iomanip>
#define endl "\n"
#define debug(x) cout << #x << " : " << x << endl;
#define debug2(x, y) cout << #x << " : " << x  << ", " << #y << " : " << y << endl;
#define FASTIO ios_base::sync_with_stdio(false),cin.tie(0);
typedef long long ll;
using namespace std;

int main()
{
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif

	FASTIO

	int t;
	cin >> t;

	while(t --)
	{
		int n;
		cin >> n;

		char a[n + 1][n + 1];
		char b[n + 1][n + 1];
		for(int i = 0; i < n; i ++)
		{
			for(int j = 0; j < n; j ++)
			{
				cin >> a[i][j];
			}
		}

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

		// Vertical Swaps
		for(int i = 0; i < n; i ++)
		{
			if(a[0][i] != b[0][i])
			{
				for(int j = 0; j < n; j ++)
				{
					if(a[j][i] == '1')
						a[j][i] = '0';
					else
						a[j][i] = '1';
				}
			}
		}

		// Horizontal Swaps
		for(int i = 0; i < n; i ++)
		{
			if(a[i][0] != b[i][0])
			{
				for(int j = 0; j < n; j ++)
				{
					if(a[i][j] == '1')
						a[i][j] = '0';
					else
						a[i][j] = '1';
				}
			}
		}

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

			if(flag == 1)
				break;
		}

		if(flag == 1)
			cout << "NO" << endl;
		else
			cout << "YES" << endl;
	}	
}