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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class SolutionF extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                                            InputStreamReader(System.in));
        }

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        new Thread(null, new SolutionF(), "Main", 1 << 26).start();
    }

    public void run() {
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            solve();
        }
        out.close();
    }

    private static void solve() {
        int n = scanner.nextInt();
        char[][] a = new char[n][n];
        for (int i = 0; i < n; i++) {
            String row =scanner.next();
            a[i] = row.toCharArray();
        }

        char[][] b = new char[n][n];
        for (int i = 0; i < n; i++) {
            String row =scanner.next();
            b[i] = row.toCharArray();
        }

        //flip first row
        boolean[] flippedColumn = new boolean[n];
        for (int i = 0; i < n; i++) {
            if (a[0][i] == b[0][i]) {
                flippedColumn[i] = true;
            }
        }
        if (isPossible(n, a, b, flippedColumn)) {
            out.println("YES");
            return;
        }

        for (int i = 0; i < n; i++) {
            flippedColumn[i] = !flippedColumn[i];
        }

        if (isPossible(n, a, b, flippedColumn)) {
            out.println("YES");
        } else {
            out.println("NO");
        }
    }

    static boolean isPossible(int n, char[][] a, char[][] b, boolean[] flippedColumn) {
        for (int r = 1; r < n; r++) {
            boolean flippedRow = false;
            boolean canFlipRow = true;
            for (int c = 0; c < n; c++) {
                if ((a[r][c] == b[r][c] && !flippedColumn[c] ||
                     a[r][c] != b[r][c] && flippedColumn[c]) && !flippedRow) {
                    canFlipRow = false;
                } else if ((a[r][c] == b[r][c] && flippedColumn[c] ||
                            a[r][c] != b[r][c] && !flippedColumn[c]) && canFlipRow) {
                    flippedRow = true;
                } else {
                    return false;
                }
            }
        }
        return true;
    }

    //REMINDERS:
    //- CHECK FOR INTEGER-OVERFLOW BEFORE SUBMITTING

    //- CAN U BRUTEFORCE OVER SOMETHING, TO MAKE IT EASIER TO CALCULATE THE SOLUTION
}