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 <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define REP(i,n) for(long long i=1; i<=n; i++)
#define rep(i,a,b) for(long long i=a; i<b; i++)
#define REPV(v) for(auto i = v.begin(); i != v.end(); ++i)
#define sort(arr) sort(arr.begin(),arr.end())
#define lb(v,n) lower_bound(v.begin(),v.end(),n)
#define ub(v,n) upper_bound(v.begin(),v.end(),n)
#define minArr(arr) *min_element(arr.begin(),arr.end())
#define maxArr(arr) *max_element(arr.begin(), arr.end())
#define findArr(arr,val) find(arr.begin(), arr.end(),val)
#define deb(a,s) cout<<a<<" "<<s<<endl
#define debV(v) for(auto x: v)printf("%lld ",x)
#define reverse(v) reverse(v.begin(),v.end())
#define printMatrix(mat) rep(i,0,mat.size()){rep(j,0,mat[i].size())cout<<mat[i][j]<<" ";cout<<endl;}
#define printMap(m) for(auto it:m)cout<<it.first<<" "<<it.second<<endl;
using namespace std;

typedef long long ll;
typedef vector<int> vii;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef unsigned long int uli;
typedef vector<vector<ll>> matrix;

const ll MOD = 1e9+7;

void xorMat(vector<vii> &matA,bool col,ll n, int index){
	if(col){
		rep(i,0,n){
			matA[i][index]^=1;
		}
	}else{
		rep(i,0,n){
			matA[index][i]^=1;
		}
	}
}

bool matchMAt(vector<vii> matA, vector<vii> matB, ll n,map<int,int> &colXor){

	rep(i,0,n)rep(j,0,n){
		if(matA[i][j] != matB[i][j]){
			if(colXor[j]==0){
				xorMat(matA,true,n,j);
				colXor[j]++;
			}else if(j==0){
				xorMat(matA,false,n,i);
			}else{
				return false;
			}
		}else{
			colXor[j]++;
		}
	}
	return true;
}

void solve() {

	ll n=1,a=1,k=1,minVal=LLONG_MAX,maxVal=LLONG_MIN,ans=LLONG_MAX,b=1,c=1,d=1;
	ll x,y,z,lx,rx,ly,ry,p,q,l,r,m,i,w;
	lx = 0,ly=0,rx=INT_MAX,ry=INT_MAX;
	cin>>n;

	vector<vii> matA(n, vii(n));
	vector<vii> matB(n, vii(n));
	map<int,int> colXor;

	string s;

	rep(i,0,n){
		cin>>s;
		rep(j,0,n)
		matA[i][j] = (int) s[j];
	}
	rep(i,0,n){
		cin>>s;
		rep(j,0,n)
		matB[i][j] = (int) s[j];
	}
	
	if(matchMAt(matA,matB,n,colXor))printf("%s\n","YES");
	else printf("%s\n","NO");
	
}

int main() {
	// your code goes here
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);	
	freopen("output.txt", "w", stdout);
#endif // ONLINE_JUDGE

	int t=1;
	string str;
	cin >> t;
	int i=1;
	//seiveFactorize(2*(1e5+1));
	while (t) {
		//cout<<"Case "<<i<<": ";
		solve();
		i++;
		t--;
	}
	return 0;
}