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.
/* Author : Anmol Gera
NSIT 4th Year */

#include<bits/stdc++.h>
using namespace std;
#define int long long


void anmol()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
#endif
}

bool isPerfectSquare(long double x) 
{   
  // Find floating point value of  
  // square root of x. 
  long double sr = sqrt(x); 
  
  // If square root is an integer 
  return ((sr - floor(sr)) == 0); 
} 

/*
while(lo < hi) {
int mid = lo + (hi - lo) / 2;
if(Special condition passed)(optional):
  return mid; 
if(condition passed)
  hi = mid;
else 
  lo = mid + 1;
}
return lo;
*/

void solve(){

//cout<<"YES"<<endl;
//cout<<"NO"<<endl;
//cout<<" "<<endl;


int n;
cin>>n;

int a[n][n];
int b[n][n];

for(int i =0; i<n; i++){

  for(int j =0; j<n; j++){
    char c ;
    cin>>c;

    a[i][j] = c- '0';

  }
}


for(int i =0; i<n; i++){

  for(int j =0; j<n; j++){
    char c ;
    cin>>c;

    b[i][j] = c- '0';

  }
}


for(int i =0; i<n; i++){

  if(a[i][0]!=b[i][0]){
    for(int j =0; j<n; j++){
      a[i][j]^=1;
    }
  }
}

for(int i =1; i<n; i++){
 if(a[0][i]!=b[0][i]){
  for(int j =0; j<n; j++){
     a[j][i]^=1;

  }
}
}

for(int i =0; i<n; i++){
  for(int j =0; j<n; j++){
    if(a[i][j]!=b[i][j]){
      cout<<"NO"<<endl;
      return;
    }
  }
}


/*int a[n];
vaector<int> v;
for(int i =0; i<n; i++){
  cin>>a[i];
}*/

//v.push_back


//cout<<"YES"<<endl;
//cout<<"NO"<<endl;
//cout<<" "<<endl;

cout<<"YES"<<endl;
 }





int32_t main(){


  anmol();
  int t;
  ///t =1;
  cin>>t;
  while(t--){


        //hhgfhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
        //hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
        //hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
        //hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
    ///  hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh'b'

        
        //int a,b;
        //cin>>a>>b;

    solve();
   //""""""""""""
        //cout<<fact[a] - fact[b];
        // look for base cases always 
       
//lkjgfdsaqwertyuiop.,mnbvcxz=-87564321`()hH" "
/*Hello Kanika,
I am Anmol Gera currently in my final year. I came across a FTE opening at Amazon - Bangalore.
The job id for the same is: 1391296
Here, is the link for the same:
https://www.amazon.jobs/en/jobs/1391296/software-development-engineer

Leetcode -https://leetcode.com/anmolgera/
Codeforces -https://codeforces.com/profile/DumbCarlsen69

I have Solved around 1000 questions on all the sites Combined. Have participated in more than 70 contests on codeforces and Currently a specialist.
You can see my cv, I have attached that as well.
Thanks a lot for your valuable time.
*/

  }

}