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.
/*  
             ____     __                       __
     /\     |  __>   /    |   |    /\    |    /
    /  \    |  \    <     |---|   /  \   |   <
   /    \   |   \    \__  |   |  /    \  |    \__

         _   __             ____    ___
        | \_/ |    /\     /        |
        |     |   /  \   <  <-->   |---
        |     |  /    \   \__\ |   |___

  Author::::::::: Archaic_Mage :::::::::Author

     _____    __________  _________     ___ ___      _____    .___  _________  
   /  _  \   \______   \ \_   ___ \   /   |   \    /  _  \   |   | \_   ___ \ 
  /  /_\  \   |       _/ /    \  \/  /    ~    \  /  /_\  \  |   | /    \  \/ 
 /    |    \  |    |   \ \     \____ \    Y    / /    |    \ |   | \     \____
 \____|__  /  |____|_  /  \______  /  \___|_  /  \____|__  / |___|  \______  /
         \/          \/          \/         \/           \/                \/ 
               _____       _____      ________  ___________                  
              /     \     /  _  \    /  _____/  \_   _____/                  
             /  \ /  \   /  /_\  \  /   \  ___   |    __)_                   
            /    Y    \ /    |    \ \    \_\  \  |        \                  
            \____|__  / \____|__  /  \______  / /_______  /                  
                    \/          \/          \/          \/                   


*/

#include <bits/stdc++.h>

typedef long long ll;
typedef int64_t i64;
typedef unsigned long long ull;

#define f(i, a, b) for (ll i= a; i < b; i++)
#define fd(i, a, b) for(ll i = a; i >= b; i--)
#define pb push_back
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pss pair<string, string>
#define pcc pair<char, char>
#define p pair
#define vpii vector<pii>
#define vi vector<int>
#define vc vector<char>
#define vll vector<ll>
#define v vector

using namespace std;

bool check(vector<vector<int>> a, vector<vector<int>> const &b) {
  int n = (int) a.size();
  for (int j = 0; j < n; j++) {
    if (a[0][j] != b[0][j]) {
      for (int i = 0; i < n; i++) {
        a[i][j] ^= 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int need_xor = (a[i][0] ^ b[i][0]);
    for (int j = 1; j < n; j++) {
      if (need_xor != (a[i][j] ^ b[i][j])) {
        return false;
      }
    }
  }
  return true;
}

void solve() {
  int n;
  cin >> n;
  vector<vector<int>> a(n, vector<int>(n));
  vector<vector<int>> b(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < n; j++) {
      a[i][j] = s[j] - '0';
    }
  }
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < n; j++) {
      b[i][j] = s[j] - '0';
    }
  }

  for (int times = 0; times < 2; times++) {
    if (check(a, b)) {
      cout << "YES\n";
      return;
    }
    for (int j = 0; j < n; j++) {
      a[0][j] ^= 1;
    }
  }
  cout << "NO\n";
}

int main() {
  int test;
  cin >> test;
  while (test-- > 0) {
    solve();
  }
  return 0;
}