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>
#define int long long
using namespace std;

void fliprow(vector<char> &row, int n)
{
    for(int j = 0; j < n; j++)
        if(row[j] == '0')
            row[j] = '1';
        else
            row[j] = '0';
}
int32_t main()
{
    ios_base::sync_with_stdio(false), cin.tie(NULL);

    int t, n;

    cin >> t;

    while (t--)
    {
        cin >> n;

        vector<vector<char>> A(n, vector<char>(n)), B(n, vector<char>(n));
        unordered_map<int,int> flipped;
        bool flag = true;
        
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                cin >> A[i][j];

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                cin >> B[i][j];

                if(B[i][j] != A[i][j])
                    A[i][j] = '1';
                else
                    A[i][j] = '0';
            }
        }

        if(A[0][0] != '1')
            fliprow(A[0], n);

        for(int j = 0; j < n; j++)
        {
            if(A[0][j] == '1')
            {
                for(int i = 1; i < n; i++)
                {
                    if(A[i][j] != '1')
                    {
                        if(!flipped[i])
                        {
                            fliprow(A[i], n);
                            flipped[i] = 1;
                        }
                        else
                        {
                            flag = false;
                            break;
                        }
                    }
                }
            }
            if(!flag)
                break;
        }

        if(flag)
        {
            for(int j = 0; j < n; j++)
            {
                int cnt1 = 0;
                for(int i = 0; i < n; i++)
                    if(A[i][j] == '1')
                        cnt1++;
                if(cnt1 > 0 && cnt1 < n)
                {
                    flag = false;
                    break;
                }
            }
        }

        if(flag)
            cout << "YES\n";
        else
            cout << "NO\n";
    }
    return 0;
}