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.Arrays;
import java.util.StringTokenizer;

public class F {
    final int INF = 0x3f3f3f3f;
    BufferedReader  br;
    PrintWriter     out;
    StringTokenizer st;
    boolean         eof;

    public F() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) {
        new F();
    }

    private void solve() {
        int t = nextInt();
        while (t-- > 0) {
            innerSolve();
        }
    }

    private void innerSolve() {
        int n = nextInt();
        int[][] matrix = new int[n][n];
        for (int i = 0; i < n; i++) {
            String s = nextString();
            byte[] bytes = s.getBytes();
            for (int j = 0, bytesLength = bytes.length; j < bytesLength; j++) {
                byte b = bytes[j];
                if (b == '0') matrix[i][j] = 0;
                else matrix[i][j] = 1;
            }
        }
        String s1 = nextString();
        int[][] matrix2 = new int[n][n];
        for (int i = 0; i < n; i++) {
            String s = nextString();
            byte[] bytes = s.getBytes();
            for (int j = 0, bytesLength = bytes.length; j < bytesLength; j++) {
                byte b = bytes[j];
                if (b == '0') matrix2[i][j] = 0;
                else matrix2[i][j] = 1;
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] != matrix2[i][j]) {
                    if (j == 0) {
                        flipHeng(matrix, i);
                    } else if (i == 0) {
                        flipShu(matrix, j);
                    } else {
                        outln("NO");
                        return;
                    }
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] != matrix2[i][j]) {
                    outln("NO");
                    return;
                }
            }
        }
        outln("YES");
    }

    private void flipShu(int[][] matrix, int i) {
        for (int j = 0; j < matrix.length; ++j) {
            matrix[j][i] = 1 - matrix[j][i];
        }
    }

    private void flipHeng(int[][] matrix, int i) {
        for (int j = 0; j < matrix.length; ++j) {
            matrix[i][j] = 1 - matrix[i][j];
        }
    }


    private void outln(Object o) {
        out.println(o);
    }

    private void out(Object o) {
        out.print(o);
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

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

    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

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

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

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


}