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.PrintWriter;
import java.util.*;

public class F {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, out);
        }
        out.close();
    }
    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        char[][] ca = new char[n][n];
        char[][] cb = new char[n][n];
        for (int i = 0; i < n; i++) {
            ca[i] = in.next().toCharArray();
        }
        for (int i = 0; i < n; i++) {
            cb[i] = in.next().toCharArray();
        }
        int[][] dp = new int[n][n];
        List<int[]> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (ca[i][j] != cb[i][j]){
                    res.add(new int[]{i, j});
                }
            }
        }
        if (res.size() == 0){
            out.println("YES");
        }else{
            int i = res.get(0)[0];
            int j = res.get(0)[1];
            if (row(i, j, n, ca, cb) && helper(i, j, n, ca, cb)){
                out.println("YES");
            }else{
                out.println("NO");
            }
        }

    }
    static boolean row(int i, int j, int n, char[][] ca, char[][] cb){
        int[] row = new int[n];
        for (int k = 0; k < n; k++) {
            if (ca[k][j] == cb[k][j]){
                row[k]++;
            }
        }
        for (int k = 0; k < n; k++) {
            char[] cp = new char[n];
            for (int l = 0; l < n; l++) {
                if (row[l] % 2 == 0){
                    cp[l] = ca[l][k];
                }else{
                    cp[l] = ca[l][k] == '0' ? '1' : '0';
                }
            }
            int cnt = 0;
            for (int l = 0; l < n; l++) {
                if (cp[l] != cb[l][k]){
                    cnt++;
                }
            }
            if (cnt != 0 && cnt != n) return false;
        }
        return true;
    }
    static boolean helper(int i, int j, int n, char[][] ca, char[][] cb){
        int[] col = new int[n];
        for (int k = 0; k < n; k++) {
            if (ca[i][k] == cb[i][k]){
                col[k]++;
            }
        }
        for (int k = 0; k < n; k++) {
            char[] cp = new char[n];
            for (int l = 0; l < n; l++) {
                if (col[l] % 2 == 0){
                    cp[l] = ca[k][l];
                }else{
                    cp[l] = ca[k][l] == '0'? '1' :'0';
                }
            }
            int cnt = 0;
            for (int l = 0; l < n; l++) {
                if (cp[l] != cb[k][l]){
                    cnt++;
                }
            }
            if (cnt != 0 && cnt != n) return false;
        }
        return true;
    }

}
