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>
typedef long long ll ;
#define int ll
const int MAX=1e3+10;
const int modn=1e9+7;
const int INF=0x3f3f3f3f3f3f;
#define endl '\n'
#define cwk     freopen("D:\\workplace\\CLion\\in.in","r",stdin),freopen("D:\\workplace\\CLion\\out.out","w",stdout)
using namespace std;
int n,m;
char aa[MAX][MAX],bb[MAX][MAX];
int a[MAX][MAX],b[MAX][MAX];
string s;
bool check_hang(int m) {
    bool is=(a[m][1]==b[m][1]);
    for (int i = 2; i <= n; i++) {
        if (is!=(a[m][i] == b[m][i])) {
            return false;
        }
    }
    return true;
}
bool check_lie(int m) {
    bool is=(a[1][m]==b[1][m]);
    for (int i = 2; i <= n; i++) {
        if (is!=(a[i][m] == b[i][m])) {
            return false;
        }
    }
    return true;
}
void solve(){
    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>aa[i][j];
            a[i][j]=aa[i][j]-'0';
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>bb[i][j];
            b[i][j]=bb[i][j]-'0';
        }
    }
    if(!check_hang(1)){
        for(int i=1;i<=n;i++){
            if(a[1][i]!=b[1][i]){
                for(int j=1;j<=n;j++){
                    a[j][i]^=1;
                }
            }
        }
        for(int i=2;i<=n;i++){
            if(!check_hang(i)){
                cout<<"NO\n";
                return;
            }
        }
        cout<<"YES\n";
        return;
    }
    bool is=true;
    for(int i=2;i<=n;i++){
        if(!check_hang(i)) is=false;
    }
    if(is){
        cout<<"YES\n";
        return;
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            a[j][i]^=1;
        }
    }
    for(int i=2;i<=n;i++){
        if(!check_hang(i)){
            cout<<"NO\n";
            return;
        }
    }
    cout<<"YES\n";

}
signed main() {
    //ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    //cwk;
    int _=1;
    cin>>_;
    while(_--){
        solve();
    }
    return 0;
}