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 <algorithm>
#include <bitset>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#define pb push_back
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mem(a) memset(a, 0, sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
typedef pair<int, int> P;
typedef __int128_t int128;
const int maxn = 1e3 + 10;
const ll mod = 1e9 + 7;
char a[maxn][maxn];
char b[maxn][maxn];
bitset<maxn> bit[maxn];
bitset<maxn> bit1;
int n;
int main()
{
    IOS;
    int t;
    cin >> t;
    bit1.set();
    while (t--)
    {
        cin>>n;
        for (int i = 1; i <= n;i++)
            cin >> a[i] + 1;
        for (int i = 1; i <= n;i++)
            cin >> b[i] + 1;
        for (int i = 1; i <= n;i++)
        {
            for (int j = 1; j <= n;j++)
            {
                bit[i][j] = (a[i][j] - '0') ^ (b[i][j] - '0');
            }
        }
        for (int i = 1; i <= n;i++)
        {
            if(bit[i][1])
                bit[i] ^= bit1;
        }
        bool f = 1;
        for (int i = 1; i <= n;i++)
        {
            int tmp1 = bit[1][i];
            for (int j = 1;j<=n;j++)
            {
                if(bit[j][i]!=tmp1)
                    f = 0;
                if(!f)
                    break;
            }
            if(!f)
                break;
        }
        cout << (f ? "YES" : "NO") << endl;
    }
    return 0;
}
 	   		  	   			  	 	  	 	 		 	