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

public class Solution1382C1 {
  public static void main(String[] args) {
    InputReader in = new InputReader(System.in);
    PrintWriter out = new PrintWriter(System.out);
    Solver1382C1 solver = new Solver1382C1();
    int n = in.nextInt();
    for (int i = 0; i < n; i++) {
      solver.solve(i, in, out);
    }
    out.close();
  }

  static class Solver1382C1 {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt();
      String a = in.next(), b = in.next();
      char[] ac = a.toCharArray();
      char[] bc = b.toCharArray();

      List<Integer> result = new ArrayList<>();
      int r = n - 1;

      while (true) {
        while (r >= 0) {
          if (ac[r] != bc[r]) {
            break;
          }
          r--;
        }

        if (r < 0) {
          break;
        }

        if (ac[0] != ac[r]) {
          result.add(1);
          ac[0] = ac[r];
        }

        result.add(r + 1);

        for (int i = 0; i <= r/2; i++) {
          if (ac[i] == ac[r-i]) {
            if (ac[i] == '1') {
              ac[i] = '0';
              ac[r-i] = '0';
            } else {
              ac[i] = '1';
              ac[r-i] = '1';
            }
          }
        }
      }

      out.print(result.size());
      for (Integer i : result) {
        out.print(" " + i);
      }
      out.println();
    }
  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }
  }
}