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

public class Main {


    public static void main(String[] args) throws IOException {
//        br = new BufferedReader(new FileReader(new File("cutting.in")));
//        out = new PrintWriter(new File("cutting.out"));
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        int PUTEN = nextInt();
        for (int POMOGI = 0; POMOGI < PUTEN; POMOGI++) {
            int n = nextInt();
            char[] a = next().toCharArray();
            char[] b = next().toCharArray();
            ArrayList<Integer> ans = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                if (a[i] != b[i]) {
                    if (i > 0) ans.add(1 + i);
                    ans.add(1);
                    if (i > 0) ans.add(1 + i);
                }
            }
            out.print(ans.size() + " ");
            for (int i = 0; i < ans.size(); i++) {
                out.print(ans.get(i) + " ");
            }
            out.println();
        }
        out.close();
    }

    static BufferedReader br;
    static PrintWriter out;
    static StringTokenizer in = new StringTokenizer("");

    static String next() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

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

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

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

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }
}
