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.
import java.util.*;
public class Main {
    public static int[][] a=new int[1005][1005];
    public static int n;
    public static int[][] b=new int[1005][1005];
    public static char[][] c=new char[1005][1005];
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            int t=scanner.nextInt();
            while (t>0){
                n=scanner.nextInt();
                for (int i=0;i<n;i++)
                    c[i]=scanner.next().toCharArray();
                for(int i=0; i<n; i++)
                    for(int j=0; j<n; j++)
                        a[i][j] = c[i][j]-'0';
                for (int i=0;i<n;i++)
                    c[i]=scanner.next().toCharArray();
                for(int i=0; i<n; i++)
                    for(int j=0; j<n; j++)
                        b[i][j] = c[i][j]-'0';
                boolean flag=true;
                for(int i=0; i<n; i++)
                    if(a[i][0] != b[i][0]){
                        for(int j=0; j<n; j++)
                            a[i][j] ^= 1;
                    }
                for(int i=0; i<n; i++) {
                    if (a[0][i] != b[0][i]) {
                        for(int j=1; j<n; j++)
                            if (a[j][i] == b[j][i]) {
                                flag = false;
                                break;
                            }
                    }
                    else if(a[0][i] == b[0][i]) {
                        for(int j=1; j<n; j++)
                            if (a[j][i] != b[j][i]) {
                                flag = false;
                                break;
                            }
                    }
                }
                if(flag)
                    System.out.println("YES");
                else
                    System.out.println("NO");
                t--;
            }
        }
    }
}
   		 		  			 		 	   	  	 		  	