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 {
        MyScanner scn = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        // Solution Start
        int t = scn.nextInt();
        while (t-- > 0) {
            solve(scn, out);
        }
        //Solution end
        out.close();
    }

    public static void solve(MyScanner scn, PrintWriter out) {
        int n = scn.nextInt();
        String a = scn.nextLine();
        String b = scn.nextLine();
        ArrayList<Integer> moves = new ArrayList<>();
        int t = n;
        for (int i = 0; i < n; i++) {
            if (a.charAt(n - i - 1) == b.charAt(n - i - 1)) {
                t--;
                continue;
            }
            if (a.charAt(0) == b.charAt(n - i - 1)) {
                a = flip(a, 1);
                moves.add(1);
            }
            a = flip(a, t);
            moves.add(t);
            t--;
        }
        out.print(moves.size() + " ");
        for (int move : moves) {
            out.print(move + " ");
        }
        out.println();

    }

    public static String flip(String a, int prefix) {
        StringBuilder newA = new StringBuilder();
        for (int i = 0; i < prefix; i++) {
            char c = a.charAt(i);
            if (c == '0') newA.append("1");
            if (c == '1') newA.append("0");
        }
        newA.reverse();
        int i = prefix;
        while (i < a.length()) {
            newA.append(a.charAt(i));
            i++;
        }

        return new String(newA);

    }

    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
    //--------------------------------------------------------
}