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;
typedef long long int ll;
const ll N=1000000007;
int a[1010][1010],b[1010][1010],c[1010][1010];

int main()
{
     ios_base::sync_with_stdio(false);
     cin.tie(NULL);
    ll  i,j,n,m,k;
    int tt=1;
    cin>>tt;
    while(tt--)
    {
        cin>>n;
        char ch;
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                cin>>ch;
                if(ch=='1')
                    a[i][j]=1;
                else
                    a[i][j]=0;
            }
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                cin>>ch;
                if(ch=='1')
                    b[i][j]=1;
                else
                    b[i][j]=0;
            }
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(a[i][j]==b[i][j])
                    c[i][j]=1;
                else
                    c[i][j]=0;
            }
        }
        int z=1;
        for(i=1;i<n;i++)
        {
            if(c[i][0]==c[i-1][0])
            {
                for(j=0;j<n;j++)
                {
                    if(c[i][j]!=c[i-1][j])
                    {
                        z=0;break;
                    }
                }
            }
            else
            {
                for(j=0;j<n;j++)
                {
                    if(c[i][j]==c[i-1][j])
                    {
                        z=0;break;
                    }
                }
            }
            if(z==0)
                break;
        }
        if(z)cout<<"YES\n";
        else cout<<"NO\n";
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                a[i][j]=b[i][j]=c[i][j]=0;

    }
    return 0;
}
