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.*;
import java.util.Map.Entry;

import java.math.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		// Scanner in = new Scanner(System.in);
		// Scanner in = new Scanner(new BufferedReader(new
		// InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(System.out);
		// InputReader in = new InputReader(new
		// File("ethan_traverses_a_tree.txt"));
		// PrintWriter out = new PrintWriter(new
		// File("ethan_traverses_a_tree-output.txt"));

		int pi = in.nextInt();
		for (int qi = 0; qi < pi; qi++) {
			int n = in.nextInt();
			int[][] a = new int[n][n];
			int[][] b = new int[n][n];

			for (int i = 0; i < n; i++) {
				char[] tmpa = in.next().toCharArray();
				for (int j = 0; j < n; j++) {
					a[i][j] = tmpa[j] - '0';
					b[i][j] = tmpa[j] - '0';
				}
			}

			int[][] dst = new int[n][n];
			for (int i = 0; i < n; i++) {
				char[] tmpa = in.next().toCharArray();
				for (int j = 0; j < n; j++) {
					dst[i][j] = tmpa[j] - '0';
				}
			}

			changeRow(n, b, 0);
			if (check(n, a, dst) || check(n, b, dst)) {
				out.printf("YES\n");
			} else {
				out.printf("NO\n");
			}
		}

		out.close();
	}

	static public void changeRow(int n, int[][] a, int row) {
		for (int j = 0; j < n; j++) {
			a[row][j] = 1 - a[row][j];
		}
	}

	static public void changeColumn(int n, int[][] a, int column) {
		for (int i = 0; i < n; i++) {
			a[i][column] = 1 - a[i][column];
		}
	}

	static public boolean check(int n, int[][] a, int[][] dst) {
		for (int j = 0; j < n; j++) {
			if (a[0][j] != dst[0][j]) {
				changeColumn(n, a, j);
			}
		}

		for (int i = 1; i < n; i++) {
			if (a[i][0] != dst[i][0]) {
				changeRow(n, a, i);
			}

			for (int j = 0; j < n; j++) {
				if (a[i][j] != dst[i][j]) {
					return false;
				}
			}
		}

		return true;
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

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

		public boolean hasNext() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

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

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}