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.math.BigInteger;
import java.util.*;


public class A {

  void solve() throws IOException {
    int t = nextInt();
    while (t-- > 0) {
      int n = nextInt();
      int[] a = new int[n];
      int[] b = new int[n];
      String s1 = nextToken();
      for (int i = 0; i < n; i++) {
        a[i] = s1.charAt(i) == '0' ? 0 : 1;
      }
      String s2 = nextToken();
      for (int i = 0; i < n; i++) {
        b[i] = s2.charAt(i) == '0' ? 0 : 1;
      }
      ArrayList<Integer> result = new ArrayList<>();
      for (int i = n - 1; i >= 0; i--) {
        if (a[i] != b[i]) {
          if (i == 0) {
            result.add(1);
          } else {
            if (a[0] == b[i]) {
              result.add(1);
              a[0] = a[0] == 0 ? 1 : 0;
            }
            result.add(i + 1);
            for (int j = 0, k = i; j <= k; j++, k--) {
              if (j == k) {
                a[j] = a[j] == 0 ? 1 : 0;
              } else {
                int tmp = a[j] == 0 ? 1 : 0;
                a[j] = a[k] == 0 ? 1 : 0;
                a[k] = tmp;
              }
            }
          }
        }
      }
      out.print(result.size());
      for (int i = 0; i < result.size(); i++) {
        out.print(" " + result.get(i));
      }
      out.println();
    }
  }

  public static void main(String[] args) throws IOException {
    new A().run();
  }

  void run() throws IOException {
    reader = new BufferedReader(new InputStreamReader(System.in));
//		reader = new BufferedReader(new FileReader("input.txt"));
    tokenizer = null;
    out = new PrintWriter(new OutputStreamWriter(System.out));
//		out = new PrintWriter(new FileWriter("output.txt"));
    solve();
    reader.close();
    out.flush();

  }

  BufferedReader reader;
  StringTokenizer tokenizer;
  PrintWriter out;

  int nextInt() throws IOException {
    return Integer.parseInt(nextToken());
  }

  long nextLong() throws IOException {
    return Long.parseLong(nextToken());
  }

  double nextDouble() throws IOException {
    return Double.parseDouble(nextToken());
  }

  String nextToken() throws IOException {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine());
    }
    return tokenizer.nextToken();
  }
}