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 <iostream>
#include <vector>
#include <array>

using namespace std;

const size_t MAX_N = 1000;

array<char, MAX_N * MAX_N> arr_a;
array<char, MAX_N * MAX_N> arr_b;
array<bool, MAX_N> col_inverted;

int main(int argc, char** argv)
{
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    size_t T;
    cin >> T;

    while(T--)
    {
        size_t N;
        cin >> N;

        col_inverted.fill(false);
        for(size_t i = 0; i < N * N; ++i)
        {
            cin >> arr_a[i];
        }

        for(size_t i = 0; i < N * N; ++i)
        {
            cin >> arr_b[i];
        }


        for(size_t i = 0; i < N; ++i)
        {
            if(arr_a[i] != arr_b[i])
            {
                col_inverted[i] = true;
            }
        }

        bool is_same = true;
        for(size_t row = 1; row < N; ++row)
        {
            const char a_first_val = col_inverted[0] ^ (arr_a[row * N] == '0') ? '0' : '1';
            const char b_first_val = arr_b[row * N];
            bool invert_row = a_first_val != b_first_val;

            for(size_t col = 1; col < N; ++col)
            {
                size_t index = row * N + col;

                const char a_val = invert_row ^ col_inverted[col] ^ (arr_a[index] == '0') ? '0' : '1';
                const char b_val = arr_b[index];

                if(a_val != b_val)
                {
                    is_same = false;
                    break;
                }

            }

            if(!is_same)
            {
                break;
            }
        }

        if(is_same)
        {
            cout << "YES\n";
        }
        else
        {
            cout << "NO\n";
        }
    }

    return 0;
}