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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author El Mehdi ASSALI
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    A2PrefixFlipHardVersion solver = new A2PrefixFlipHardVersion();
    int testCount = Integer.parseInt(in.next());
    for (int i = 1; i <= testCount; i++)
      solver.solve(i, in, out);
    out.close();
  }

  static class A2PrefixFlipHardVersion {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt();
      String s1 = in.next(), s2 = in.next();
      List<Integer> perms = new ArrayList<>();
      int l = 0, r = n - 1, xor = 0;
      for (int i = n - 1; i >= 0; i--) {
        char c1 = xor == 0 ? s1.charAt(l) : s1.charAt(r);
        int d1 = (c1 - '0') ^ xor;
        int d2 = s2.charAt(i) - '0';
        if (d1 == d2) {
          perms.add(1);
        }
        perms.add(i + 1);
        if (xor == 0) {
          l++;
        } else {
          r--;
        }
        xor ^= 1;
      }
      out.print(perms.size());
      perms.forEach(e -> out.print(" " + e));
      out.println();
    }

  }

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

    public InputReader(InputStream inputStream) {
      reader = new BufferedReader(new InputStreamReader(inputStream));
    }

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

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

  }
}