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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

/**
 * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
 */

public class Main {
   public static void main(String[] args) {
      // Write your solution here
      FastReader in = new FastReader();
      int tt = in.nextInt();
      while(tt-- > 0){
         int n = in.nextInt();
         String s = in.nextLine();
         String t = in.nextLine();
         List<Integer> l1 = new ArrayList<>();
         List<Integer> l2 = new ArrayList<>();
         s += '0';
         t += '0';
         for(int i = 1; i <= n; i++){
            if(s.charAt(i) != s.charAt(i - 1)){
               l1.add(i);
            }
            if(t.charAt(i) != t.charAt(i - 1)){
               l2.add(i);
            }
         }
         Collections.reverse(l2);
         l1.addAll(l2);
         System.out.print(l1.size() + " ");
         for(int num : l1){
            System.out.print(num + " ");
         }
         System.out.println();
      }
   }
   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;
      }
   }
}