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.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class prefix {
            static String reverse(String s){

              char r[]=s.toCharArray();
              for(int i=0;i<r.length;i++){
                if(r[i]=='1'){
                  r[i]='0';
                }else{
                  r[i]='1';
                }
              }
              String p="";
              for(int i=r.length;i>=0;i--){
                p+=r[i];
              }
              return p;
            }
          
    public static void main(String[] args) throws IOException {
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
      
          while(t-->0){
            int n=sc.nextInt();
            StringBuilder sb=new StringBuilder();
            int ac=0;
            String a=sc.next();
            String b=sc.next();
            for(int i=n-1;i>=0;i--){
              if(a.charAt(i)!=b.charAt(i)){
                ac+=3;
                  sb.append((i+1)+" "+1+" "+(i+1)+" ");
              }
            }
            System.out.print(ac+" ");
            System.out.println(sb.toString());
            

































          }
        

    }
}
