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 Main {

    private static void run(Reader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int[][] a = new int[n][n];
        int[][] b = new int[n][n];
        for (int i = 0; i < n; i++) {
            char[] c = in.next().toCharArray();
            for (int j = 0; j < n; j++) {
                a[i][j] = c[j] - '0';
            }
        }
        in.nextLine();
        for (int i = 0; i < n; i++) {
            char[] c = in.next().toCharArray();
            for (int j = 0; j < n; j++) {
                b[i][j] = c[j] - '0';
            }
        }

        if (n == 1) {
            out.println("yes");
            return;
        }

        boolean flag1 = true;
        boolean flag2 = true;
        int[] change1 = new int[n];
        int[] change2 = new int[n];
        for (int i = 0; i < n; i++) {
            change1[i] = a[0][i] != b[0][i] ? 1 : 0;
            change2[i] = a[0][i] == b[0][i] ? 1 : 0;
        }

        for (int i = 1; i < n; i++) {
            int line_change_1 = 0;
            int line_change_2 = 0;
            for (int j = 0; j < n; j++) {
                int now_1 = a[i][j] ^ change1[j] ^ line_change_1;
                int now_2 = a[i][j] ^ change2[j] ^ line_change_2;

                if (now_1 != b[i][j]) {
                    if (j == 0) line_change_1 = 1;
                    else flag1 = false;
                }

                if (now_2 != b[i][j]) {
                    if (j == 0) line_change_2 = 1;
                    else flag2 = false;
                }
            }
        }
        if (flag1 || flag2) {
            out.println("yes");
        } else {
            out.println("no");
        }
    }

    public static void main(String[] args) throws IOException {
        Reader in = new Reader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

        int t = in.nextInt();
        for (int i = 0; i < t; i++) {
            run(in, out);
        }

        out.flush();
        in.close();
        out.close();
    }

    static class Reader {
        BufferedReader reader;
        StringTokenizer st;

        Reader(InputStreamReader stream) {
            reader = new BufferedReader(stream, 32768);
            st = null;
        }

        void close() throws IOException {
            reader.close();
        }

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

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

        String nextLine() throws IOException {
            return reader.readLine();
        }

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

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

    }
}