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;
import java.util.Arrays;
import java.util.StringTokenizer;

public class F {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int inputs = Integer.parseInt(in.readLine());
		while(inputs-->0) {
			StringTokenizer st = new StringTokenizer(in.readLine());
			int n = Integer.parseInt(st.nextToken());
			char[][] mat = new char[n][n];
			for(int r = 0; r < n; r++)
				mat[r] = in.readLine().toCharArray();
			char[][] wanted = new char[n][n];
			in.readLine();
			for(int r = 0; r < n; r++)
				wanted[r] = in.readLine().toCharArray();
			boolean[][] bad = new boolean[n][n];
			for(int r = 0; r < n; r++) {
				for(int c = 0; c < n; c++) {
					if(wanted[r][c] == mat[r][c])
						bad[r][c] = false;
					else
						bad[r][c] = true;
				}
			}
			boolean pos1 = true;
			boolean[] flipped = new boolean[n];
			for(int c = 0; c < n; c++) {
				if(bad[0][c])
					flipped[c] = true;
			}
			for(int r = 1; r < n; r++) {
				boolean flip = bad[r][0]^flipped[0];
				for(int c = 1; c < n; c++) {
					if(bad[r][c]^flipped[c]^flip) {
						pos1 = false;
						break;
					}
				}
			}
			boolean pos2 = true;
			Arrays.fill(flipped, false);
			for(int c = 0; c < n; c++) {
				if(!bad[0][c])
					flipped[c] = true;
			}
			for(int r = 1; r < n; r++) {
				boolean flip = bad[r][0]^flipped[0];
				for(int c = 1; c < n; c++) {
					if(bad[r][c]^flipped[c]^flip) {
						pos2 = false;
						break;
					}
				}
			}
			if(pos1 || pos2)
				System.out.println("YES");
			else
				System.out.println("NO");
		}

	}

}
