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: Daniel
 * Created Time: 2021-01-26 10:41:50
**/

// time-limit: 2000
#include <bits/stdc++.h>

using namespace std;

#define F first 
#define S second
#define ER erase
#define IS insert
#define PI acos(-1)
#define PB pop_back
#define EB emplace_back
#define lowbit(x) (x & -x)
#define SZ(x) ((int)x.size())
#define MP(x, y) make_pair(x, y)
#define ALL(x) x.begin(), x.end()
#define RALL(x) x.rbegin(), x.rend()
#define SOS; ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout<<fixed<<setprecision(10);

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef unsigned long long ULL;

template <typename A> using VE = vector<A>;
template <typename A> using USET = unordered_set<A>;
template <typename A> using HEAP = priority_queue<A>;
template <typename A, typename B> using PA = pair<A, B>;
template <typename A, typename B> using UMAP = unordered_map<A, B>;
template <typename A> using RHEAP = priority_queue<A, vector<A>, greater<A> >;

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////

// check the limitation!!!
const int N = 100010, M = 1010;

int T;
int n;

// read the question carefully!!!
int main()
{
    SOS;

    cin >> T;
    while (T -- )
    {
        cin >> n;
        VE<string> a(n), b(n);
        for (int i = 0; i < n; i ++ ) cin >> a[i];
        for (int i = 0; i < n; i ++ ) cin >> b[i];

        for (int i = 0; i < n; i ++ )
        {
            if (a[i][0] == '1')
                for (int j = 0; j < n; j ++ )
                    a[i][j] ^= '0' ^ '1';
            if (b[i][0] == '1')
                for (int j = 0; j < n; j ++ )
                    b[i][j] ^= '0' ^ '1';
        }

        for (int j = 0; j < n; j ++ )
        {
            if (a[0][j] == '1')
                for (int i = 0; i < n; i ++ )
                    a[i][j] ^= '0' ^ '1';
            if (b[0][j] == '1')
                for (int i = 0; i < n; i ++ )
                    b[i][j] ^= '0' ^ '1';
        }

        cout << (a == b ? "YES" : "NO") << '\n';
    }
    return 0;
}

// GOOD LUCK!!!
