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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class F {
    public static void main(String[] args) {
        new F().solve(System.in, System.out);
    }

    public void solve(InputStream in, OutputStream out) {
        InputReader reader = new InputReader(in);
        PrintWriter writer = new PrintWriter(new BufferedOutputStream(out));

        int t = reader.nextInt();
        for (int t1 = 0; t1 < t; t1++) {
            int n = reader.nextInt();
            List<String> src = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                src.add(reader.next());
            }
            List<String> target = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                target.add(reader.next());
            }
            boolean result = solve(n, src, target);
            writer.println(result ? "YES" : "NO");
        }

        writer.close();
    }

    public boolean solve(int n, List<String> srcS, List<String> targetS) {
        int[][] a = parse(srcS);
        int[][] b = parse(targetS);
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < n; j++) {
                int first = get(a, i, j);
                int second = get(b, i, j);
                if ((first % 2) != (second % 2)) {
                    return false;
                }
            }
        }
        return true;
    }

    private int[][] parse(List<String> in) {
        int[][] result = new int[in.size()][in.size()];
        for (int i = 0; i < in.size(); i++) {
            for (int j = 0; j < in.size(); j++) {
                result[i][j] = (in.get(i).charAt(j) == '1') ? 1 : 0;
            }
        }
        return result;
    }

    private int get(int[][] in, int i, int j) {
        return in[0][0] + in[i][0] + in[0][j] + in[i][j];
    }

    private static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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