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 LL long long
using namespace std;
int T,n,f[2004],r[2005];
char a[1001][1001],b[1001][1001];
int find(int x)
{
    int t=f[x];
    if(f[x]!=x)
    {
        f[x]=find(f[x]);
        r[x]=(r[x]+r[t]+2)%2;
    }
    return f[x];
}
bool judge(int x,int y)
{
    int xx=find(x),yy=find(y);
    return xx==yy;
}
void uni(int x,int y, int ra)
{
    int xx=find(x),yy=find(y);
    f[xx]=yy;
    r[xx]=(ra+r[y]-r[x]+2)%2;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n*2;i++)f[i]=i,r[i]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                a[i][j]=getchar();
                while(a[i][j]!='1'&&a[i][j]!='0')a[i][j]=getchar();
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                b[i][j]=getchar();
                while(b[i][j]!='1'&&b[i][j]!='0')b[i][j]=getchar();
            }
        }
        int fl=1;
        for(int i=1;i<=n;i++)
        {
            if(!fl)break;
            for(int j=1;j<=n;j++)
            {
                int x=i,y=j+n;
                if(a[i][j]==b[i][j])
                {
                    if(!judge(x,y))uni(x,y,0);
                    else
                    {
                        int xx=find(x),yy=find(y);
                        int d=r[x]+r[y];
                        d%=2;
                        if(d!=0){fl=0;break;}
                    }
                    
                }
                else
                {
                    if(!judge(x,y))uni(x,y,1);
                    else
                    {
                        int xx=find(x),yy=find(y);
                        int d=(r[x]+r[y])%2;
                        if(d!=1){fl=0;break;}
                    }
                }
            }
        }
        printf(fl?"YES\n":"NO\n");
    }
}