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>

#define ld long double
#define ll long long
using namespace std;
template<class T>
void read(T& x)
{
	T res = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {
		if (c == '-')f = -1; c = getchar();
	}
	while (isdigit(c)) {
		res = (res << 3) + (res << 1) + c - '0'; c = getchar();
	}
	x = res * f;
}
const ll N = 2000 + 10;
const int mod = 1e9 + 7;
int f[N * 10];
int t, n, a[N][N], b[N][N];
int find(int x)
{
	return x == f[x] ? f[x] : f[x] = find(f[x]);
}
#define r(x) (x+n)
#define c(x) (x)
#define fm(x) (x+2*n)
int main()
{
	//ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
#endif // ONLINE_JUDGE
	read(t);
	while (t--)
	{
		read(n);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%1d", &a[i][j]);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%1d", &b[i][j]);
		for (int i = 1; i <= n; i++)
		{
			f[r(i)] = r(i);
			f[fm(r(i))] = fm(r(i));
			f[c(i)] = c(i);
			f[fm(c(i))] = fm(c(i));
		}
		bool isok = 1;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (a[i][j] == b[i][j])
				{
					if (find(r(i)) == find(fm(c(j))) || find(c(j)) == find(fm(r(i))))isok = 0;
					f[find(r(i))] = find(c(j));
					f[find(fm(r(i)))] = find(fm(c(j)));
				}
				else
				{
					if (find(r(i)) == find((c(j))) || find(fm(c(j))) == find(fm(r(i))))isok = 0;
					f[find(r(i))] = find(fm(c(j)));
					f[find(fm(r(i)))] = find(c(j));
				}
			}
		}
		printf("%s\n", isok ? "YES" : "NO");
	}

	return 0;
}
