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.
//ceil(a/b) is equal to (a+b-1)/b
#include<bits/stdc++.h>
 
#define mx(a) *max_element(all(a))
#define mn(a) *min_element(all(a))
#define int long long int
 
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define vec vector <int>
#define pr vector<pair<int,int>> 
#define vecv vector <vec>
#define vecs vector <string>
#define vecc vector <char>
 
#define pi 3.141592653589793238
#define mod 1000000007
#define Mod 998244353
#define inf 1e16
#define precise(i) fixed << setprecision(10) << i
#define endl "\n"
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define one cout<<-1<<endl;

#define err1(a) cout<<#a<<" "<<a<<endl;
#define err2(a,b) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<endl;
#define err3(a,b,c) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<endl;
 
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); 
#define all(a) a.begin(),a.end()
#define show(a) for(auto xyz:a)cout<<xyz<<" ";cout<<endl;
#define show_nl(a) for(auto xyz:a)cout<<xyz<<endl;
#define loop(a) for(auto xyz:a)
#define f(i,a,b) for(int i=a;i<b;i++)
#define rf(i,a,b) for(int i=b;i>=a;i--)

const long double eps = 1e-8;

#define lb lower_bound  //THIS GIVES THE ITR TO THE ELEMENT IN RANGE [SI,EI) THAT IS >= OUR VALUE 
#define ub upper_bound  //THIS GIVES THE ITR TO THE ELEMENT IN RANGE [SI,EI) THAT IS > OUR VALUE 
 
using namespace std;
 
signed main() {
     fastio
     int t=1;
     cin>>t;
     while(t--)
     {
         int n;
        cin>>n;
        vecs s1(n),s2(n);
        for(auto &i:s1)
        cin>>i;
        for(auto &i:s2)
        cin>>i;
        f(i,0,n)
        {
            if(s1[0][i]!=s2[0][i])
            {
                f(j,0,n)
                {
                    if(s1[j][i]=='0') s1[j][i]='1';
                    else s1[j][i]='0';
                }
            }
        }
        f(i,0,n)
        {
            if(s1[i][0]!=s2[i][0])
            {
                f(j,0,n)
                {
                    if(s1[i][j]=='0') s1[i][j]='1';
                    else s1[i][j]='0';
                }
            }
        }
        int f1=0;
        f(i,0,n)
        {
            f(j,0,n) if(s1[i][j]!=s2[i][j]) f1=1;
        }
        if(f1) no 
        else yes
     }
}