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.
//NEVER GIVE UP ON YOUR DREAMS 

import java.io.*;
import java.util.*;

public class TaskF {
    private static final int MOD = (int) 1e9 + 7;
    private static final int INF = (int) 1e8;
    private static int MAXN = (int) 1e5 + 6;
    private static FastReader fr = new FastReader();

    public static void main(String[] args) {
        StringBuilder sb = new StringBuilder();
        //Code here
        int tt=fr.nextInt();
        while(tt-->0) {
            int n = fr.nextInt();
            char a[][] = new char[n][n];
            char b[][] = new char[n][n];
            for (int i = 0; i < n; i++) {
                a[i] = fr.next().toCharArray();
            }
            for (int i = 0; i < n; i++) {
                b[i] = fr.next().toCharArray();
            }
            if(a[0][0]!=b[0][0]){
                flipRow(a,0);
            }
            for(int i=1;i<n;i++){
                if(a[0][i]!=b[0][i]){
                    flipCol(a,i);
                }
            }
            for(int i=1;i<n;i++){
                if(a[i][0]!=b[i][0]){
                    flipRow(a,i);
                }
            }
            //check
            boolean ok=true;
            outer: for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if(a[i][j]!=b[i][j]){
                        ok=false;
                        break outer;
                    }
                }
            }
            sb.append(ok?"YES\n":"NO\n");
        }
        System.out.println(sb);
    }
    static void flipRow(char a[][],int r){
        for(int i=0;i<a.length;i++){
            a[r][i]=(a[r][i]=='1'?'0':'1');
        }
    }
    static void flipCol(char a[][],int r){
        for(int i=0;i<a.length;i++){
            a[i][r]=(a[i][r]=='1'?'0':'1');
        }
    }
    //FAST IO here
    static int[] readArray(int n) {
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = fr.nextInt();
        }
        return a;
    }

    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 ie) {
                    ie.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;
        }
    }
}