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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class F {

	public static void main(String[] args) {
		Scanner sc =new Scanner(System.in);
		int t =sc.nextInt();
		for(;t>0;t--) {
			int n =sc.nextInt();
			sc.nextLine();
			int a[][]=new int[n][n];
			for(int i = 0; i<n;i++) {
				String s = sc.nextLine();
				for(int j = 0;j<n;j++) {
					a[i][j]=Integer.parseInt(s.charAt(j)+"");
				}
			}
			sc.nextLine();
			
			int b[][]=new int[n][n];
			for(int i = 0; i<n;i++) {
				String s = sc.nextLine();
				for(int j = 0;j<n;j++) {
					b[i][j]=Integer.parseInt(s.charAt(j)+"");
				}
			}
			//System.out.println("balabezo");
			
			boolean row[]=new boolean[n];
			boolean column[]=new boolean[n];
			for(int i = 0;i<n;i++) {
				if(a[0][i]!=b[0][i]) {
					column[i]=true;
				}
			}
			for(int i =1;i<n;i++) {
				if(a[i][0]!=b[i][0]&&!column[0]) {
					row[i]=true;
				}
				else if(a[i][0]==b[i][0]&&column[0]) {
					row[i]=true;
				}
			}
			boolean f = true;
           // System.out.println(Arrays.toString(column));
           // System.out.println(Arrays.toString(row));
			for(int i = 0;i<n&&f;i++) {
				for(int j = 0;j<n&&f;j++) {
					toggel(i, j, a, row, column);
					if(a[i][j]!=b[i][j])f=false;
				}
			}
			//pri(a);
			//pri(b);
			if(f)System.out.println("YES");
			else System.out.println("NO");

		}

	}
	
	public static void toggel(int i,int j,int[][]a,boolean [] row , boolean [] column) {
		
		if(row[i]) {
			//System.out.println(i+" "+j);
		if(a[i][j]==0)a[i][j]=1;
		else a[i][j]=0;
		}
		if(column[j]) {
			//System.out.println(i+" "+j);
			if(a[i][j]==0)a[i][j]=1;
			else a[i][j]=0;	
		}
	}
	public static void pri(int a[][]) {
		for(int i= 0;i<a.length;i++) {
			System.out.println(Arrays.toString(a[i]));
		}
	}
	
	static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {

            return Long.parseLong(next());
        }
    }


}
