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 Main {
  public static void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

    int T = Integer.parseInt(in.readLine());
    for (int t = 0; t < T; ++t) {
      int n = Integer.parseInt(in.readLine());
      String a = in.readLine();
      String b = in.readLine();

      List<Integer> result = solve(n, a, b);
      System.out.print(result.size() + " ");
      for (int p : result) {
        System.out.print(p + " ");
      }
      System.out.println();
    }

    in.close();
  }

  private static List<Integer> solve(int n, String a, String b) {
    List<Integer> result = new ArrayList<>();

    char[] chars = a.toCharArray();
    for (int i = n - 1; i >= 0; --i) {
      if (b.charAt(i) != chars[i]) {
        if (b.charAt(i) == chars[0]) {
          result.add(1);
          chars[0] = reverse(chars[0]);
        }

        result.add(i + 1);
        reverse(chars, i + 1);
      }
    }

    return result;
  }

  private static void reverse(char[] chars, int n) {
    for (int i = 0; i < n / 2; ++i) {
      char temp = chars[i];
      chars[i] = reverse(chars[n - 1 - i]);
      chars[n - 1 - i] = reverse(temp);
    }

    if (n % 2 == 1) {
      chars[n / 2] = reverse(chars[n / 2]);
    }
  }

  private static char reverse(char ch) {
    return (char) ('0' + ('1' - ch));
  }
}