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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;

public class F {
	static ArrayList<Integer> primes;
	static HashSet<Integer> primesH;
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		PrintWriter out=new PrintWriter(System.out);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			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]='1'-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]='1'-s.charAt(j);
				}
			}
			if(a[0][0]!=b[0][0]) {
				for(int i=0;i<n;i++) {
					a[i][0]^=1;
				}
			}
			for(int i=1;i<n;i++) {
				if(a[0][i]!=b[0][i]) {
					for(int j=0;j<n;j++) {
						a[j][i]^=1;
					}
				}
			}
			for(int i=1;i<n;i++) {
				if(a[i][0]!=b[i][0]) {
					for(int j=0;j<n;j++) {
						a[i][j]^=1;
					}
				}
			}
			boolean f=true;
			for(int i=0;i<n&&f;i++) {
				for(int j=0;j<n;j++) {
					if(a[i][j]!=b[i][j]) {
						f=false;break;
					}
				}
			}
			out.println(f?"YES":"NO");
			
		}
		
		out.close();

	}
	static int gcd (int a, int b) {
	    return b==0?a:gcd (b, a % b);
	}
	static boolean isPrime(long n) {
		if(n<1000000) {
			return primesH.contains((int)n);
		}
		for(int p: primes)
			if(p * p > n)
				break;
			else if(n % p == 0)
				return false;
		return true;
	}
	static void sieveOfEratosthenes(int n)  // O(n*log(log(n)))
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        boolean prime[] = new boolean[n+1]; 
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
          
        // Print all prime numbers 
        for(int i = 2; i <= n; i++) 
        { 
            if(prime[i] == true) {
               // System.out.print(i + " "); 
                primes.add(i);
                primesH.add(i);
            }
        } 
    } 
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public boolean hasNext() {return st.hasMoreTokens();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		
		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}
			
		public boolean ready() throws IOException {return br.ready(); }
		public int[] nextArrInt(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		public long[] nextArrLong(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

	}

}
