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.*;

public class Practice {

	public static void main(String[] args) throws Exception {
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t-->0) {
			int n=s.nextInt();
			int[][]pre=new int[n][n];
			int[][]fin=new int[n][n];
			int[]h=new int[n];
			int[]v=new int[n];
			for(int i=0;i<n;i++) {
				String str=s.next();
				for(int j=0;j<n;j++) {
					pre[i][j]=Integer.valueOf(str.charAt(j)+"");
					//System.out.print(pre[i][j]+" ");
				}//System.out.println();
			}
			for(int i=0;i<n;i++) {
				String str=s.next();
				for(int j=0;j<n;j++) {
					fin[i][j]=Integer.valueOf(str.charAt(j)+"");
					///System.out.print(fin[i][j]);
				}//System.out.println();
			}
			
			if(pre[0][0]!=fin[0][0]) {
				h[0]=1;
				if(getAns(pre,fin,h,v)) {
					System.out.println("YES");
					continue;
				}
				h=new int[n];
				v=new int[n];
				
				v[0]=1;
				if(getAns(pre,fin,h,v)) {
					System.out.println("YES");
					continue;
				}
				
			}else {
				if(getAns(pre,fin,h,v)) {
					System.out.println("YES");
					continue;
				}
				h=new int[n];
				v=new int[n];
				
				h[0]=1;
				v[0]=1;
				if(getAns(pre,fin,h,v)) {
					System.out.println("YES");
					continue;
				}
			}
			System.out.println("NO");
		}
	    
	}

	public static boolean getAns(int[][] pre, int[][] fin, int[] h, int[] v) {
		// TODO Auto-generated method stub
		for(int i=1;i<h.length;i++) {
			if(pre[0][i]!=fin[0][i]) {
				if(v[0]%2!=1) {
					h[i]=1;
				}
			}else {
				if(v[0]%2==1) {
					h[i]=1;
				}
			}
			
			if(pre[i][0]!=fin[i][0]) {
				if(h[0]%2==1) {
					
				}else {
					v[i]=1;
				}
			}else {
				if(h[0]%2==1) {
					v[i]=1;
				}else {
					
				}
			}
		}
		
		for(int i=1;i<h.length;i++) {
			for(int j=1;j<h.length;j++) {
				//System.out.println(i+" "+j+" "+(v[i]+h[j]));
				if((v[i]+h[j])%2==0) {
					//System.out.println("OOOO");
					//System.out.println(pre[i][j]+" "+fin[i][j]);
					if(pre[i][j]!=fin[i][j]) {
						return false;
					}
				}else {
					if(pre[i][j]==fin[i][j]) {
						return false;
					}
				}
			}
		}
		return true;
	}
}