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
#define MAX 100002
#define mod 1000000007
#define INF 1000000007
#define pb push_back
#define ff first
#define ss second
#define w(tt) int tt;cin>>tt;while(tt--)
#define endl  "\n"
#define IOS  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

void solve(){
	 int n;
	 cin>>n;
	 char x;
	 vector<vector<int>> to(n,vector<int>(n)),frm(n,vector<int>(n));
	 for(int i=0;i<n;i++){
		 for(int j=0;j<n;j++){cin>>x;frm[i][j]=(x-'0');}
		 }
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){cin>>x;to[i][j]=(x-'0');
			}
		}
	for(int i=0;i<n;i++){
		if(frm[0][i]!=to[0][i]){
			for(int j=0;j<n;j++)frm[j][i]^=1;
			}
		}
	int f=0;
	for(int i=1;i<n;i++){
		vector<int> v1,v0,v;
		for(int j=0;j<n;j++){
			v1.pb(frm[i][j]^1);
			v0.pb(frm[i][j]);
		}
		for(int j=0;j<n;j++)v.pb(to[i][j]);
		if(v0!=v && v1!=v)f=1;
		}
		if(f==0)cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
int32_t main(){
	IOS;
	w(tt){solve();}
	//solve();
	return 0;
	}








