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.io.*;

import java.util.*;



public class problemA 
{
   public static void main (String[] args) throws IOException
   {
   
      BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer token = new StringTokenizer(input.readLine());
      int maxCount = Integer.parseInt(token.nextToken());
      for(int b = 0; b < maxCount; b++)
      {
         token = new StringTokenizer(input.readLine());
         int maxLength = Integer.parseInt(token.nextToken());
         String first = input.readLine();
         String second = input.readLine();
         int count = 0;
         String operations = "";
         for(int x = maxLength - 1; x >= 0; x--)
         {
            if(Character.getNumericValue(first.charAt(x)) != Character.getNumericValue(second.charAt(x)))
            {
               if(Character.getNumericValue(first.charAt(0)) != Character.getNumericValue(second.charAt(x)))
               {
                  count++;
                  operations = operations + (x + 1) + " ";
                  first = prefixSwap(first, x);
               }
               else
               {
                  count = count + 2;
                  operations = operations + 1  + " " + (x + 1) + " ";
                  first = prefixSwap(first, 0);
                  first = prefixSwap(first, x);
               }
            }
         }
         operations = operations.trim();
         System.out.println(count + " " + operations);
      }     
   }
   public static String prefixSwap(String s, int index)
   {
      String answ = "";
      for(int x = 0; x <= index; x++)
      {
         if(Character.getNumericValue(s.charAt(x)) == 1)
         {
            answ = "0" + answ;
         }
         else
         {
            answ = "1" + answ;
         }
      }
      if(index == s.length() - 1)
         return answ;
      else
         return answ + s.substring(index + 1);
   }
}