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<iostream>
#include<vector>
#include<string>
#include<utility>
#include<algorithm>
#define loop(x,i,n) for(int x = i; x < n; ++ x)
#define looprev(x,i,n) for(int x = i;x>=n;x--)
using namespace std;
typedef long long ll;
vector <int> memo;
vector <bool> out;
int dp(int);
int main()
{
    ll size;
    cin>>size;
    while(size-->0)
    {
        int n;
        cin>>n;
        //cout<<"n is "<<n<<"\n";
        if(n==1)
        {
            cin>>n>>n;
            out.push_back(true);
            continue;
        }
        string temp ="";
        vector<vector<int> >M (n);
        loop(i,0,n)
        {
            cin>>temp;
            //cout<<"temp is "<<temp<<"\n";
            loop(j,0,n)
                M[i].push_back((temp.at(j)=='0')?0:1);
        }

        loop(i,0,n)
        {
            cin>>temp;
            //cout<<"temp is "<<temp<<"\n";
            loop(j,0,n)
            {
                int x = (temp.at(j)=='0')?0:1;
                M[i][j] = M[i][j] ^ x;
            }
        }

        vector<int> XOR = M[0];
        bool flag = true;
        loop(i,1,n)
        {
            loop(j,1,n)
            {
                if(M[i][j] ^ M[i][j-1] != XOR[j] ^ XOR[j-1])
                {
                    flag = false;
                    break;
                }
            }
            if(!flag)
                break;
        }
        out.push_back(flag);
    }
    for(bool i: out)
    {
        cout<<(i?"YES":"NO")<<"\n";
    }

    return 0;
}