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 dd double
#define ll long long int 
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
ll mod=1e9+7;
bool func(vector<vector<char> > &v1,vector<vector<char> > &v2){
    vector<ll> rem(v1.size(),0),rem1(v1.size(),0);
    for (int i = 0; i < v1.size(); i++)
    {
       vector<char> temp1=v2[i];
       vector<char>temp2(v1.size());
       for(int j=0;j<v1.size();j++){
           if(rem[j]%2!=0){
              temp2[j]=(v1[i][j]=='1')?'0':'1';
           }
           else temp2[j]=v1[i][j];
       }
       if(temp1!=temp2){
          int t;
          for( t=0;t<temp1.size();t++){
            if(temp1[t]==temp2[t]) break;
          }
          if(t==temp1.size()) rem1[i]+=1;
          else{
             for(int t=0;t<temp1.size();t++){
                 if(temp1[t]!=temp2[t]) rem[t]+=1;
             }
          }
       }
    }
    for(int i=0;i<rem.size();i++){
       if(rem[i]%2!=0){
          for(int j=0;j<v1.size();j++){
              v1[j][i]=(v1[j][i]=='1')?'0':'1';
          }
       }
    }
    for(int i=0;i<rem1.size();i++){
      if(rem1[i]%2!=0){
         for(int j=0;j<v1[0].size();j++){
              v1[i][j]=(v1[i][j]=='1')?'0':'1';
         }
      }
    }
    return v1==v2;
}
int main()
{
  #ifndef ONLINE_JUDGE
  freopen("input1.txt","r",stdin);
  freopen("output.txt","w",stdout);
  #endif
  fastio;
  ll test_case;
  test_case=1;
  cin>>test_case;
  while(test_case--){
       ll n;
       cin>>n;
       vector<vector<char> >v1(n,vector<char>(n)),v2(n,vector<char>(n));
       for (int i = 0; i < n; ++i)
       {
        for (int j = 0; j < n; ++j)
        {
            cin>>v1[i][j];
        }
       }
       for (int i = 0; i < n; ++i)
       {
        for (int j = 0; j < n; ++j)
        {
            cin>>v2[i][j];
        }
       }
       bool ans=func(v1,v2);
       (ans)?cout<<"YES"<<endl:cout<<"NO"<<endl;
  }
}