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 _ ios_base::sync_with_stdio(0); cin.tie(0);
#define endl '\n'
#define INF 0x3f3f3f3f
#define MAX (1<<10)
#define MOD 1000000007
#define i64 long long
#define all(x) (x).begin() , (x).end()
#define sz(x) (int)(x).size()
#define ii pair<int, int>
#define fs first
#define sc second
#define eb emplace_back
#define vi vector<int>
#define vvi vector<vi>
#define vii vector<ii>
#define vvii vector<vii>
#define lsb(x) ((x) & (-x))
#define gcd(x,y) __gcd((x),(y))
#define W(x) cerr << "\033[31m"<<  #x << " = " << x << "\033[0m" << endl;

using namespace std;

int t, n, m, k;
int a[MAX][MAX], b[MAX][MAX];
string s;
int ans;
int cj[2*MAX];

int main(){_
    for (cin >> t; t--;){
        cin >> n;
        for (int i=0; i<n; ++i){
            cin >> s;
            for (int j=0; j<n; ++j){
                a[i][j] = s[j]-'0';
            }
        }

        for (int i=0; i<n; ++i){
            cin >> s;
            for (int j=0; j<n; ++j){
                a[i][j] ^= (s[j]-'0');
            }
        }

        cj[0] = 0;
        for (int i=0; i<n; ++i){
            cj[i+n] = a[0][i];
        }

        ans = 1;
        for (int i=1; i<n; ++i){
            cj[i] = (cj[n]^a[i][0]);
            for (int j=1; j<n; ++j){
                int x = i, y = j+n;
                int d = (a[i][j]);

                ans &= (cj[x] == (cj[y]^d));
            }
        }
        cout << (ans ? "YES" : "NO") << endl;
    }
    return 0;
}
