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.StringTokenizer;

public class codeforces697_F {
    private static void solve(FastIOAdapter ioAdapter) {
        int n = ioAdapter.nextInt();

        int[][] a = new int[n][];
        int[][] a2 = new int[n][];
        for (int i = 0; i < n; i++) {
            int[] cur = new int[n];
            int[] cur2 = new int[n];
            char[] chars = ioAdapter.next().toCharArray();
            for (int j = 0; j < n; j++) {
                if (chars[j] == '0') {
                    cur[j] = 0;
                    cur2[j] = 0;
                } else {
                    cur[j] = 1;
                    cur2[j] = 1;
                }
            }
            a[i] = cur;
            a2[i] = cur2;
        }
        int[][] b = new int[n][];
        for (int i = 0; i < n; i++) {
            int[] cur = new int[n];
            char[] chars = ioAdapter.next().toCharArray();
            for (int j = 0; j < n; j++) {
                if (chars[j] == '0') {
                    cur[j] = 0;
                } else {
                    cur[j] = 1;
                }
            }
            b[i] = cur;
        }

        if (check(a, b)) {
            ioAdapter.out.println("YES");
        } else {
            for (int i = 0; i < n; i++) {
                a2[0][i] ^= 1;
            }

            if (check(a2, b)) {
                ioAdapter.out.println("YES");
            } else {
                ioAdapter.out.println("NO");
            }
        }
    }

    private static boolean check(int[][] a, int[][] b) {
        for (int i = 0; i < a.length; i++) {
            if (a[0][i] != b[0][i]) {
                for (int j = 0; j < b.length; j++) {
                    a[j][i] ^= 1;
                }
            }
        }

        for (int i = 0; i < a.length; i++) {
            int needXor = a[i][0] ^ b[i][0];
            for (int j = 1; j < b.length; j++) {
                if (needXor != (a[i][j] ^ b[i][j])) {
                    return false;
                }
            }
        }
        return true;
    }

    public static void main(String[] args) throws Exception {
        try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
            int count = 1;
            count = ioAdapter.nextInt();
            while (count-- > 0) {
                solve(ioAdapter);
            }
        }
    }

    static class FastIOAdapter implements AutoCloseable {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
        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());
        }

        @Override
        public void close() throws Exception {
            out.flush();
            out.close();
            br.close();
        }
    }
}