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"
#include<algorithm>
#include<iostream>
#include<ctime>
#include<cstdlib>
#include<vector>
#include<cassert>
#include<climits>
#include<ctime>
#include<math.h>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define ff first
#define ss second
#define FOR(i, j, k, in) for (int i=j ; i<k ; i+=in)
#define rep(i, j) FOR(i, 0,     j, 1)
#define MP make_pair
#define PB push_back
#define INF (int)1e9+1
#define endl "\n"
#define MOD 1000000007
#define MOD1  998244353
#define DEB(x) cout<<"## "<<x<<" ##"<<endl;
typedef pair<int,int> pii;
const int SIZE=(int)2e5+5;
//Change point of view
//Don't get stuck
//Easier solution exists 
int par[SIZE];
void build(int n){
    for(int i=1;i<=n;i++){
        par[i]=i;
    }
}
int find(int x){
    if(par[x]==x){
        return x;
    }
    return par[x]=find(par[x]);
}
bool set_union(int x,int y){
    x=find(x);
    y=find(y);
    if(x!=y){
        par[x]=y;
        return true;
    }
    return false;
}
int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int t;
    cin>>t;
    while(t--){
        int n,flag=0;
        cin>>n;
        string a[n],b[n];
        rep(i,n){
            cin>>a[i];
        }
        rep(i,n){
            cin>>b[i];
        }
        for(int i=0;i<n;i++){
            if(a[0][i]!=b[0][i]){
                rep(j,n){
                    a[j][i]^=1;
                }
            }
        }
        for(int i=0;i<n;i++){
            if(a[i][0]!=b[i][0]){
                rep(j,n){
                    a[i][j]^=1;
                }
            }
        }
        rep(i,n){
            rep(j,n){
                if(a[i][j]!=b[i][j]){
                    flag=-1;
                }
            }
        }
        if(flag==0){
            cout<<"YES";
        }
        else{
            cout<<"NO";
        }
        cout<<endl;
    }
    return 0;
}