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 pcarp{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
			int n = sc.nextInt();
			int[][] arr1 = new int[n][n];
			int[][] arr2 = new int[n][n];
			for(int i=0;i<n;i++){
				String s = sc.next();
				for(int j=0;j<s.length();j++){
					arr1[i][j]=s.charAt(j)-'0';
				}
			}
			//String s1 = sc.next();
			for(int i=0;i<n;i++){
				String s = sc.next();
				for(int j=0;j<s.length();j++){
					arr2[i][j]=s.charAt(j)-'0';
				}
			}
			for(int i=0;i<n;i++){
				if(arr1[0][i]!=arr2[0][i]){
					for(int j=0;j<n;j++){
						arr1[j][i]^=1;
					}
				}
			}
			for(int i=0;i<n;i++){
				if(arr1[i][0]!=arr2[i][0]){
					for(int j=0;j<n;j++){
						arr1[i][j]^=1;
					}
				}
			}
			int flag=0;
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
						if(arr1[i][j]!=arr2[i][j]){
							flag=1;
							break;
						}
				}
			}
			if(flag==0){
				System.out.println("YES");
			}
			else{
				System.out.println("NO");
			}
		}
	}
	public static long fact( long n, long mod){
		long res =1;
		for(int i=1;i<=n;i++){
			res%=mod;
			i%=mod;
			res=(res*i)%mod;
		}
		return res;
	}

	public static long nCk(long n,long k, long mod){
		return (fact(n,mod)%mod*modular(fact(k,mod),mod-2,mod)%mod*modular(fact(n-k,mod),mod-2,mod)%mod)%mod;
	}

	public static long modular(long n, long e, long mod){
		long res = 1;
		n%=mod;
		if (n == 0) return 0;
		while(e>0){
			if((e&1)==1){
				res=(res*n)%mod;
			}
			e=e>>1;
			n=(n*n)%mod;
		}
		return res;
	}

}