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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class unusualmatrix {

    static boolean[][] finish;
    static boolean[][] cur;
    static int n;


    static void flipfirst(int row){
        for (int i = 0; i < n; i++) {
            cur[row][i] = !cur[row][i];
        }
    }

    static void flipsecond(int row){
        for (int i = 0; i < n; i++) {
            cur[i][row] = !cur[i][row];
        }
    }

    static boolean check(){
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if(cur[i][j] != finish[i][j]){
                    return false;
                }
            }
        }
        return true;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        for (int ff = 0; ff < t; ff++) {
            n = Integer.parseInt(br.readLine());
            cur = new boolean[n][n];
            for (int i = 0; i < n; i++) {
                String s = br.readLine();
//                System.out.println("S: " + s);
                for (int j = 0; j < n; j++) {
                    cur[i][j] = s.charAt(j) == '1';
                }
            }
            br.readLine();
            finish = new boolean[n][n];
            for (int i = 0; i < n; i++) {
                String s= br.readLine();
                for (int j = 0; j < n; j++) {
                    finish[i][j] = s.charAt(j) == '1';
                }
            }
            //try without flipping any columns=
            if(check()){
                System.out.println("YES");
                continue;
            }
            //try only flipping the inside n-1 x n-1 matrix
            boolean[][] backup = new boolean[n][n];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    backup[i][j] = cur[i][j];
                }
            }
            for (int i = 0; i < n; i++) {
                if(cur[i][0] != finish[i][0])
                    flipfirst(i);
            }
            for (int i = 0; i < n; i++) {
                if(cur[0][i] != finish[0][i])
                    flipsecond(i);
            }
            if(check()){
                System.out.println("YES");
                continue;
            }
            System.out.println("NO");
        }
    }
}
