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<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<vector>
#include<sstream>
#include<algorithm>
#include<bitset>
#include<stack>
#include<map>
#include<set>
#include<ctime>
#include<cstdlib>
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define mst(a,b) memset(a,b,sizeof(a))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define dbg(a) cout<<"* "<< #a <<" : "<<a<<endl
#define fr first
#define se second
//#define ls x<<1
//#define rs x<<1|1
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define lowbit(x) x&(-x)
#define ac cout<<ans<<endl
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-7;
const ull hashp=131;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn=2e3+100;
char s[maxn];
int g[maxn][maxn],n;
int check(int x,int y){
	if(!g[x][y]){
		if(x<n&&y<n&&!g[x+1][y]&&!g[x][y+1]&&g[x+1][y+1])return 0;
		if(x>1&&y<n&&!g[x-1][y]&&!g[x][y+1]&&g[x-1][y+1])return 0;
		if(x<n&&y>1&&!g[x+1][y]&&!g[x][y-1]&&g[x+1][y-1])return 0;
		if(x>1&&y>1&&!g[x-1][y]&&!g[x][y-1]&&g[x-1][y-1])return 0;
	}
	else {
		if(g[x+1][y]&&g[x][y+1]&&!g[x+1][y+1])return 0;
		if(g[x-1][y]&&g[x][y+1]&&!g[x-1][y+1])return 0;
		if(g[x+1][y]&&g[x][y-1]&&!g[x+1][y-1])return 0;
		if(g[x-1][y]&&g[x][y-1]&&!g[x-1][y-1])return 0;
	}
	return 1;
}
void solve(){
	int _=1;scd(_);
	//init();
	for(;_--;){
		scd(n);
		rep1(i,n+1)rep1(j,n+1)g[i][j]=0;
		rep(___,2)rep1(i,n){
			scanf("%s",s+1);
			rep1(j,n)g[i][j]^=(s[j]-'0');
		}
		int ans=1;
		rep1(i,n)rep1(j,n){
			ans&=check(i,j);
		}
		if(ans)puts("YES");
		else puts("NO");
	}
}
int main(){
	//IOS;
	//freopen("D:\\data.txt","r",stdin);
	//freopen("D:\\out.txt","w",stdout);
    solve();
    return 0;
}
