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 rep(i, l, r) for(int i = (l), i##end = (r); i <= i##end; i++)
#define per(i, r, l) for(int i = (r), i##end = (l); i >= i##end; i--)
#define mem(x, val) memset(x, val, sizeof(x))
#define debug(x) cerr<<#x<<" = "<<x
#define pii pair<int, int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
using namespace std;
typedef long long ll;
const int N = 5000+10, inf = 0x3f3f3f3f;
vector<int> edge[N];
int T, n, tot, cnt, a[N][N], dfn[N], low[N], vis[N], color[N];
stack<int> st;
void tarjan(int u)
{
	dfn[u] = low[u] = ++tot;
	st.push(u);
	vis[u] = 1;
	rep(i, 0, edge[u].size()-1)
	{
		int v = edge[u][i];
		if(!dfn[v])
		{
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else
			if(vis[v])
				low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u])
	{
		cnt++;
		int x;
		do
		{
			x = st.top();
			st.pop();
			color[x] = cnt;
			vis[x] = 0;
		} while(x != u);
	}
}
int main()
{
	scanf("%d", &T);
	while(T--)
	{
		tot = cnt = 0;
		mem(dfn, 0);
		mem(vis, 0);
		mem(low, 0);
		mem(color, 0);
		while(st.size())
			st.pop();
		scanf("%d", &n);
		rep(i, 1, 4*n)
			edge[i].clear();
		rep(i, 1, n)
			rep(j, 1, n)
				scanf("%1d", &a[i][j]);
		rep(i, 1, n)
		{
			rep(j, 1, n)
			{
				int x;
				scanf("%1d", &x);
				if(a[i][j]^x)
				{
					edge[i*2-1].pb(j*2+2*n);
					edge[j*2-1+2*n].pb(i*2);
					edge[i*2].pb(j*2-1+2*n);
					edge[j*2+2*n].pb(i*2-1);
				}
				else
				{
					edge[i*2-1].pb(j*2-1+2*n);
					edge[j*2+2*n].pb(i*2);
					edge[j*2-1+2*n].pb(i*2-1);
					edge[i*2].pb(j*2+2*n);
				}
			}
		}
		rep(i, 1, n*4)
			if(!dfn[i])
				tarjan(i);
		rep(i, 1, n*2)
			if(color[2*i-1] == color[2*i])
			{
				printf("NO\n");
				goto l;
			}
		printf("YES\n");
		l:;
	}
	return 0;
}
    				 		  		 	   						