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.
// Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;






// Please name your class Main
public class Main {
    static Scanner in = new Scanner(System.in);
    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);

        int T=Int();
        for(int t=0;t<T;t++){
            int n=Int();
            int A[][]=new int[n][n];
            int B[][]=new int[n][n];

            for(int i=0;i<n;i++){
                String s=Str();
                for(int j=0;j<n;j++){
                    A[i][j]=s.charAt(j)-'0';
                }
            }

            for(int i=0;i<n;i++){
                String s=Str();
                for(int j=0;j<n;j++){
                    B[i][j]=s.charAt(j)-'0';
                }
            }

            Solution sol=new Solution();
            sol.solution(A,B);

        }
        out.flush();
    }

    public static long Long(){
        return in.nextLong();
    }
    public static int Int(){
        return in.nextInt();
    }
    public static String Str(){
        return in.next();
    }
}


class Solution{
    //horizontal : 1-n
    //vertical   : n+1 - 2n
    List<Integer>adjecent[];
    List<Integer>rev[];
    boolean visit[];
    int order[];
    int cnt=0;
    int comp[];

    public void solution(int A[][],int B[][]){
        //0 xor 1= 1
        //1 xor 1 =0
        //=>toggle
        int n=A.length;
        int M=2*n;
        visit=new boolean[M*2+100];
        adjecent=new ArrayList[M*2+100];
        rev=new ArrayList[M*2+100];
        order=new int[M*2+1];
        comp=new int[M*2+1];
        Arrays.fill(comp,-1);

        for(int i=0;i<rev.length;i++){
            adjecent[i]=new ArrayList<>();
            rev[i]=new ArrayList<>();
        }



        //construct graph
        for(int i=0;i<n;i++){
            int u=i;
            for(int j=0;j<n;j++){
                int v=j+n;

                if(A[i][j]==B[i][j]){//0 zero toggle
                    add(u,v);
                    add(v,u);
                    add(u+M,v+M);
                    add(v+M,u+M);
                }
                else{//1 toggle
                    add(u+M,v);
                    add(v,u+M);
                    add(v+M,u);
                    add(u,v+M);
                }
            }
        }



        for(int i=0;i<2*M;i++){
            if(!visit[i]){
                dfs1(i);
            }
        }

        int color=0;
        for(int i=2*M-1;i>=0;i--){
            int v=order[i];
            if(comp[v]==-1){
                dfs2(v,color);
                color++;
            }
        }


        for(int i=0;i<M;i++){
            if(comp[i]==comp[i+M]){
                System.out.println("NO");
                return;
            }
        }

        System.out.println("YES");

    }

    public void add(int u,int v){
        adjecent[u].add(v);
        rev[v].add(u);
    }

    public void dfs1(int root){
        visit[root]=true;
        List<Integer>childs=adjecent[root];
        for(int c:childs){
            if(visit[c])continue;
            dfs1(c);
        }
        order[cnt++]=root;
    }

    public void dfs2(int root,int color){
        comp[root]=color;
        List<Integer>childs=rev[root];
        for(int c:childs){
            if(comp[c]!=-1)continue;
            dfs2(c,color);
        }
    }

}