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
#define x first
#define y second
#define PII pair<int,int>
#define LL long long
#define pb push_back
#define pqp priority_queue<PII,vector<PII>,greater<PII>>
#define pqi priority_queue<int,vector<int>,greater<int>>
#define PSS pair<string,string>
#define in insert
#define line inline
#define sort(s) sort(s.begin(),s.end());
#define reverse(s) reverse(s.begin(),s.end());
#define V vector<int>
#define VV vector<V>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;

// struct ios
// {
//     inline char read()
//     {
//         static const int IN_LEN = 1e6 + 10;
//         static char buf[IN_LEN], *s, *t;
//         return (s == t) && (t = (s = buf) + fread(buf, 1, IN_LEN, stdin)), s == t ? -1 : *s++;
//     }
//     template <typename _Tp>
//     inline ios &operator>>(_Tp &x)
//     {
//         static char c11, boo;
//         for (c11 = read(), boo = 0; !isdigit(c11); c11 = read())
//         {
//             if (c11 == -1)
//                 return *this;
//             boo |= c11 == '-';
//         }
//         for (x = 0; isdigit(c11); c11 = read())
//             x = x * 10 + (c11 ^ '0');
//         boo && (x = -x);
//         return *this;
//     }
// } io;

int dx[]=  {0,-1,0,1,0,0},dy[]= {-1,0,1,0,0,0},dz[] = {0,0,0,0,-1,1};
const int N = 1010;
char a[N][N],b[N][N];
int n;
void solve()
{
    cin >> 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] != b[i][0])
            for(int j = 0; j < n; j ++)
                a[i][j] ^= 1;

    for(int j = 1 ; j < n; j ++)
        if(a[0][j] != b[0][j]){
            for(int i = 0; i < n; i ++)
                if(a[i][j] == b[i][j]){
                    cout << "NO" << endl;
                    return ;
                }
        }
        else{
            for(int i = 0; i < n; i ++)
                if(a[i][j] != b[i][j]){
                    cout << "NO" << endl;
                    return ;
                }
        }
    cout << "YES" << endl;
}

signed main()
{
    IOS
    int _; cin >> _; while(_ --)
    solve();
}
