This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.





/*input
5 5 2 3



*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
 {
 static Scanner in = new Scanner(System.in);
 
 
    public static void solve()
    {   
        int n=in.nextInt();
        in.nextLine();
        String s=in.nextLine();
        String s1=in.nextLine();
        StringBuilder sb = new StringBuilder();
        int cnt=0;
        if(s.charAt(0)=='1')
               cnt=1;
               int tot=0;
        for(int i=n-1;i>=0;i--)
        {
           if(s1.charAt(i)!=s.charAt(i))
           {
               tot+=3;
               sb.append(i+1).append(" ");sb.append(1).append(" ");
               sb.append(i+1).append(" ");
               
               
           }
        }
        
       
        System.out.println(tot+" " +sb);
        
    }
    
	public static void main (String[] args)
	 {
	 int t=in.nextInt();
	 while(t-->0)
	 {
	  solve();
	 }
}
}












