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.*;
public class Main
{
    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  e) 
                { 
                    e.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; 
        } 
    }
    public static long modPow(long a,int p,long mod)
    {
        long res=1L;
        while(p>0)
        {
            if(p%2==1)
            {
                res=res*a;
                res%=mod;
                p--;
            }
            else
            {
                a*=a;
                a%=mod;
                p>>=1;
            }
        }
        return res;
    }
    public static void main(String args[])
    {
        FastReader fs=new FastReader();
        PrintWriter pw=new PrintWriter(System.out);
        final long mod=(long)1e9+7;
        long f[]=new long[1001];
        long inv[]=new long[1001];
        f[0]=1L;
        for(int i=1;i<=1000;i++)
        f[i]=(i*f[i-1])%mod;
        inv[1000]=modPow(f[1000],(int)1e9+5,mod);
        for(int i=999;i>=0;i--)
        inv[i]=(inv[i+1]*(i+1))%mod;
        int tc=fs.nextInt();
        while(tc-->0)
        {
            int n=fs.nextInt();
            char a[][]=new char[n][n];
            char b[][]=new char[n][n];
            for(int i=0;i<n;i++)
            a[i]=fs.nextLine().toCharArray();
            fs.nextLine();
            for(int i=0;i<n;i++)
            b[i]=fs.nextLine().toCharArray();
            int xor[]=new int[n];
            int rxor[]=new int[n];
            for(int i=0;i<n;i++)
            {
                boolean needed=false;
                for(int j=0;j<n;j++)
                {
                    int t1=(a[i][j]-'0')^xor[j];
                    int t2=(b[i][j]-'0');
                    if(t1!=t2)
                    {
                        rxor[i]=1;
                        needed=true;
                    }
                }
                for(int j=0;j<n;j++)
                {
                    int t1=(a[i][j]-'0')^xor[j];
                    int t2=(b[i][j]-'0');
                    if(t1==t2&&needed)
                    xor[j]^=1;
                }
            }
            boolean ok=true;
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    int t1=(a[i][j]-'0')^xor[j]^rxor[i];
                    int t2=(b[i][j]-'0');
                    if(t1!=t2)
                    {
                        ok=false;
                        break;
                    }
                }
                if(!ok)
                break;
            }
            pw.println(ok?"YES":"NO");
        }
        pw.flush();
        pw.close();
    }
}