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 javax.print.DocFlavor;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.nio.file.Paths;
import java.util.*;
public class Main extends PrintWriter {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));Main () { super(System.out); }public static void main(String[] args) throws IOException{ Main  d1=new Main   ();d1.main();d1.flush(); }

    void main() throws IOException {
        StringBuilder sb = new StringBuilder();
        int t=1;
        t=i(s()[0]);
        while(t-->0) {
            String[] s1=s();
           int n=i(s1[0]);
           char[][] a=new char[n][n];
           char[][] b=new char[n][n];
           for(int i=0;i<n;i++){
               a[i]=s()[0].toCharArray();
           }s();
            for(int i=0;i<n;i++){
                b[i]=s()[0].toCharArray();
            }
            int[][] c=new int[n][n];
            for(int i=0;i<n;i++){
             for(int j=0;j<n;j++){
                 if(a[i][j]!=b[i][j]) c[i][j]=1;
             }
            }int f=0;
            for(int i=1;i<n;i++){
                int flag=0;
                if(c[i][0]!=c[i-1][0]) flag=1;
                for(int j=0;j<n;j++){
                    if(Math.abs(c[i][j]-c[i-1][j])!=flag){
                        f=1;break;
                    }
                }
            }if(f==0){
                sb.append("YES\n");
            }else sb.append("NO\n");
        }
        System.out.println(sb);
    }

    static String[] s() throws IOException { return s.readLine().trim().split("\\s+"); }
    static int i(String ss) {return Integer.parseInt(ss); }
    static long l(String ss) {return Long.parseLong(ss); }
    public void arr(int[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=i(s2[i]); }}
}
class Pair{
    int a,b;double c;
    public Pair(int a,int b,double c){
        this.a=a;this.b=b;this.c=c;
    }
}