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.*;import java.io.*;import java.math.*;
public class Main
{
    public static void process()throws IOException
    {
        int n=ni();
        int[][]a=new int[n][n];
        int c1=0,c2=0;
        for(int i=0;i<n;i++)
        {
            String s=nln();
            for(int j=0;j<n;j++)
                if(s.charAt(j)=='1')
                    a[i][j]=1;
        }
        String t=nln();
        int[][]b=new int[n][n];
        for(int i=0;i<n;i++)
        {
            String s=nln();
            for(int j=0;j<n;j++)
                if(s.charAt(j)=='1')
                    b[i][j]=1;
        }
        for(int i=0;i<n;i++)
        {
            if(a[0][i]!=b[0][i])
            {
                for(int j=0;j<n;j++)
                    a[j][i]^=1;
            }
        }
        for(int i=0;i<n;i++)
        {
            int c=0;
            for(int j=0;j<n;j++)
                if(a[i][j]!=b[i][j])
                    c++;
            if(c>0 && c<n)
                {pn("NO");return;}
        }
        pn("YES");
    }

    static AnotherReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);}
        else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");}
        int t=1;
        t=ni();
        while(t-->0) {process();}
        out.flush();out.close();  
    }

    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static void pni(Object o){out.println(o);out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;}
    static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}

/////////////////////////////////////////////////////////////////////////////////////////////////////////

    static class AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br=new BufferedReader(new InputStreamReader(System.in));}
    AnotherReader(int a)throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}
   
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}