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.*; 
import java.util.*; 
import java.math.*; 

public class Main { 
	static boolean check(char[][] a,char[][] b,int[] row,int[] col) {

		int n=a.length;
		for(int i=0;i<n;i++) {
			for(int j=0;j<n;j++) {
				if((row[i]+col[j])%2==0 && a[i][j]!=b[i][j]) {
					return false;
				}
				else if((row[i]+col[j])%2==1 && a[i][j]==b[i][j]) {
					return false;
				}
			}
		}
		return true;
	}
	
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=f.nextInt();
//		int t=1;
			
		while(t>0) {
			t--;
			int n=f.nextInt();
			char[][] a=new char[n][n];
			char[][] b=new char[n][n];
			for(int i=0;i<n;i++) {
				char[] curr=f.next().toCharArray();
				for(int j=0;j<n;j++) {
					a[i][j]=curr[j];
				}
			}
			for(int i=0;i<n;i++) {
				char[] curr=f.next().toCharArray();
				for(int j=0;j<n;j++) {
					b[i][j]=curr[j];
				}
			}
			int[] row=new int[n];
			int[] col=new int[n];
			for(int i=0;i<n;i++) {
				if(a[i][0]!=b[i][0]) {
					row[i]=1;
				}
			}
			if(row[0]==0) {
				for(int i=0;i<n;i++) {
					if(a[0][i]!=b[0][i]) {
						col[i]=1;
					}
				}
			}
			else {
				for(int i=0;i<n;i++) {
					if(a[0][i]==b[0][i]) {
						col[i]=1;
					}
				}
			}
			boolean first=check(a,b,row,col);
			row=new int[n];
			col=new int[n];
			col[0]=1;
			for(int i=0;i<n;i++) {
				if(a[i][0]==b[i][0]) {
					row[i]=1;
				}
			}
			if(row[0]==0) {
				for(int i=0;i<n;i++) {
					if(a[0][i]==b[0][i]) {
						col[i]=1;
					}
				}
			}
			else {
				for(int i=0;i<n;i++) {
					if(a[0][i]!=b[0][i]) {
						col[i]=1;
					}
				}
			}
			boolean second=check(a,b,row,col);
			
			if(first || second) {
				System.out.println("YES");
			}
			else {
				System.out.println("NO");
			}
			
		}
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a .length; i++) a[i] = q.get(i);
    }
	static void sort(long[] a) {
        ArrayList<Long> q = new ArrayList<>();
        for (long i: a) q.add(i);
        Collections.sort(q);
        Collections.reverse(q);
//        System.out.println(q);
        for (int i = 0; i < a .length; i++) a[i] = q.get(i);
    }
	
	static long gcd(long a, long b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 