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;
// A function to print all prime  
// factors of a given number n 
int main() {
    
   // SieveOfEratosthenes(1000001);
   ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
long long int t,i=0;

cin>>t;
    
    
//   t=1;
    while(t--)
    {
   long long int j;
     long long int f=0,p=0,z,o=100000000,count=0,ans=0;
       double pie=3.141592653589793238;
    long long int c=100000000000,n,m,l=1,r=0,k,min1=10000000000000,max2=-100;
    vector<pair<long long int,long long int>>v;
    char ch;
    cin>>n;
   int a[n][n],b[n][n];
   for(i=0;i<n;i++)
   {
       for(j=0;j<n;j++)
    {   cin>>ch;
       a[i][j]=ch-48;
   }
   }
   
   for(i=0;i<n;i++)
   {
       for(j=0;j<n;j++)
    {   cin>>ch;
       b[i][j]=ch-48;
   }
   }
  // for(i=0;i<n;i++)
   //cout<<a[0][i]<<" ";
   for(i=0;i<n;i++)
   {
       if(a[0][i]!=b[0][i])
       {
           for(j=0;j<n;j++)
           {
               a[j][i]=(a[j][i]^1);
           }
       }
   }
   for(i=1;i<n;i++)
   {
       for(j=0;j<n;j++)
       {
           if(a[i][j]!=b[i][j]&&j==0)
           {
               for(k=0;k<n;k++)
               {
                   a[i][k]=(a[i][k]^1);
               }
           }
           else if(a[i][j]!=b[i][j])
           {
               break;
           }
           
       }
       if(j<n)
       break;
   }
   if(i<n)
   cout<<"NO"<<"\n";
   else
   cout<<"YES"<<"\n";
   
   }   
    
 return 0;
}
