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>
using namespace std;

#define int long long
#define ll long long
#define rep(i,n) for (int i = 0; i < (n); i++)  
#define rrep(i,n) for (int i = (n)-1; i >= 0; i--)
#define rap(i,a,n) for (int i = a; i < (n); i++) 
#define rrap(i,n,a) for (int i = (n)-1; i >= a; i--) 
#define LL_MAX 9223372036854775807

typedef vector<int> vi;  
typedef pair<int,int> ii;  
typedef vector<ii> vii;  
typedef list<int> li;  
typedef unordered_map<int,int> mii;

vector<vi> v1,v2;

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;

    rep(tt,t) {
        int n;
        cin >> n;

        v1.assign(n,vi());
        v2.assign(n,vi());

        rep(i,n) {
            string s;
            cin >> s;
            rep(j,n) {
                v1[i].push_back(s[j]-48);
            }
        }

        rep(i,n) {
            string s;
            cin >> s;
            rep(j,n) {
                v2[i].push_back(s[j]-48);
            }
        }

        rep(i,n) {
            if (v1[0][i] == v2[0][i]) continue;
            rep(j,n) {
                if (v1[j][i] == 0) v1[j][i] = 1;
                else v1[j][i] = 0;
            }
        }
        
        rap(i,1,n) {
            if (v1[i][0] != v2[i][0]) {
                rep(j,n) {
                    if (v1[i][j] == 0) v1[i][j] = 1;
                    else v1[i][j] = 0;
                }
            }
        }

        bool flag=false;
        rep(i,n) {
            rep(j,n) {
                if (v1[i][j] != v2[i][j]) {
                    flag=true;
                    break;
                }
            }
            if (flag) break;
        }

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

    return 0;
}