Problem: 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.

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    static int l, r;
    static char oppose(char t){
        if (t == '0') return'1';
        return '0';
    }

    static char HolderValue(int index, String s){
        if ((s.length()- index) %2 ==1 ) {
            r -- ;
            return oppose(s.charAt(r+1));
        }
        else{
            l++;
            return s.charAt(l-1);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        int t = sc.nextInt();
        String s,s1;
        char holder;
        ArrayList<Integer> res;
        int n ;
        for (int x= 0 ; x < t ; x++){
             n = sc.nextInt();
             sc.nextLine();
             s  =sc.nextLine();
             s1 = sc.nextLine();
             holder = s.charAt(0);

             res = new ArrayList<>();

             l = 1;
             r = s.length()-1 ;

             for (int i = s.length()-1; i >= 0; i--)
             if ( i == 0 ){
                 if (holder != s1.charAt(i))  res.add(1);
             }
             else
             {
                  if (holder == s1.charAt(i))
                        res.add(1);
                        res.add (i+1);

                  holder = HolderValue (i,s);
             }

             System.out.print (res.size()+" ");
             for (int i =0 ; i < res.size(); i++)
                 System.out.print (res.get(i)+" ");
             System.out.println ();
        }
       //-----------------
    }
}