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 Problem697F {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		for (int i = 0; i < t; i++) {
			int n = s.nextInt();
			int before[][] = new int[n][n];
			int after[][] = new int[n][n];
			for (int j = 0; j < n; j++) {
				String str = s.next();
				for (int k = 0; k < n; k++) {
					if (str.charAt(k) == '1') {
						before[j][k] = 1;
					} else {
						before[j][k] = 0;
					}
				}

			}
			for (int j = 0; j < n; j++) {
				String str = s.next();
				for (int k = 0; k < n; k++) {
					if (str.charAt(k) == '1') {
						after[j][k] = 1;
					} else {
						after[j][k] = 0;
					}
				}

			}
			int newMatrix[][] = new int[n][n];
			for (int j = 0; j < n; j++) {
				for (int k = 0; k < n; k++) {
					newMatrix[j][k] = before[j][k] ^ after[j][k];
					
				}
				
			}
			

			boolean op1 = true;
			boolean op2 = true;
			for (int j = 1; j < n; j++) {

				for (int k = 0; k < n; k++) {
					
					if (newMatrix[j][k] != newMatrix[0][k]) {
						op1 = false;
						break;
					}
				}
				for (int k = 0; k < n; k++) {
					
					if (newMatrix[j][k] == newMatrix[0][k]) {
						op2 = false;
						break;
					}
				}
				if(op1==false&&op2==false) {
					
					break;
				}
				op1=true;
				op2=true;
			}
			System.out.println((op1 || op2) ? "YES" : "NO");

		}
	}

}
