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:
//package Codeforce.div2.div2_658;

import java.io.*;
import java.util.*;

/**
 * @author Tran Anh Tai
 * @template for CP codes
 */
public class TaskC {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task {
        public void solve(InputReader in, PrintWriter out) {
            int test = in.nextInt();
            for (int i = 0; i < test; i++){
                int n = in.nextInt();
                String a = in.nextToken();
                String b = in.nextToken();
                int cnt = 0;
                ArrayList<Integer> indexes = new ArrayList<>();
                ArrayList<Integer> aToAllOnes = transform(a, n);
                ArrayList<Integer> bToAllOnes = transform(b, n);
                indexes.addAll(aToAllOnes);
                for (int j = bToAllOnes.size() - 1; j >= 0; j--){
                    indexes.add(bToAllOnes.get(j));
                }
                out.print(indexes.size() + " ");
                for (int j : indexes){
                    out.print(j + " ");
                }
                out.println();
            }
        }

        private ArrayList<Integer> transform(String a, int n) {
            ArrayList<Integer> ops = new ArrayList<>();
            for (int i = 1; i < n; i++){
                if (a.charAt(i) != a.charAt(i - 1)){
                    ops.add(i);
                }
            }
            if (a.charAt(n - 1) == '0'){
                ops.add(n);
            }
            return ops;
        }
    }
    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}