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.util.*;

@SuppressWarnings("unused")
public class Main {

  private static void solve() {
    int t = ni();
    for (int i = 0; i < t; i++) {
      int n = ni();
      int[][] a = new int[n][n];
      for (int j = 0; j < n; j++) {
        char[] s = ns();
        for (int k = 0; k < n; k++) {
          a[j][k] = s[k] - '0';
        }
      }
      int[][] b = new int[n][n];
      for (int j = 0; j < n; j++) {
        char[] s = ns();
        for (int k = 0; k < n; k++) {
          b[j][k] = s[k] - '0';
        }
      }
      String ret = solve(n, a, b);
      out.println(ret);
    }
  }

  private static String solve(int n, int[][] a, int[][] b) {
    for (int j = 0; j < n; j++) {
      if (a[0][j] != b[0][j]) {
        for (int i = 0; i < n; i++) {
          a[i][j] ^= 1;
        }
      }
    }

    for (int i = 0; i < n; i++) {
      int diff = 0;
      for (int j = 0; j < n; j++) {
        if (a[i][j] != b[i][j]) {
          diff++;
        }
      }
      if (diff != 0 && diff != n)
        return "NO";
    }
    return "YES";
  }

  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

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

  private static double nd() {
    return Double.parseDouble(next());
  }

  private static long nl() {
    return Long.parseLong(next());
  }

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

  private static int ni() {
    return Integer.parseInt(next());
  }

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}
