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;
#define int long long //delete if causing problems
#define F first
#define dbg(x)  cout<<#x<<" "<<x<<endl;
#define S second
#define setbit(n) __builtin_popcount(n)
#define all(x) x.begin() , x.end()
#define clr(x) memset(x,0,sizeof(x))
#define fast  ios_base::sync_with_stdio(0); cin.tie(0);
#define endl "\n" //delete if interactive
#define MOD 1000000007
const int inf = 1e18;
int power(int a, int b);
int a[1001][1001];
int b[1001][1001];


signed   main()
{
	fast
	int t = 1;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;

		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				char c;
				cin >> c;
				int x = c - '0';
				a[i][j] = x;

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

			}
		}


		int flag = 0;

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


						for (int k = 1; k <= n; k++) a[k][j] = (a[k][j] ^ 1);

					}
				}
				else {
					if (a[i][j] != b[i][j])
					{
						if (j != 1)
						{
							flag = 1;
							break;
						}

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

				}

			}

			if (flag) break;
		}



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





	}
	return 0;
}

int power(int a, int b)
{
	int res = 1;
	while (b)
	{
		if (b % 2) b-- , res = res * a;
		else b = b / 2 , a *= a;
	}
	return res;
}

/*
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
*/



