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.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        for (int tc = 0; tc < t; ++tc) {
            sc.nextInt();
            String a = sc.next();
            String b = sc.next();

            System.out.println(solve(a, b));
        }

        sc.close();
    }

    static String solve(String a, String b) {
        List<Integer> operations = new ArrayList<>();
        char[] bits = a.toCharArray();
        for (int i = bits.length - 1; i >= 0; --i) {
            if (bits[i] != b.charAt(i)) {
                if (i != 0 && bits[0] == b.charAt(i)) {
                    operations.add(1);
                    operate(bits, 1);
                }

                operations.add(i + 1);
                operate(bits, i + 1);
            }
        }

        return String.format("%d %s", operations.size(),
                operations.stream().map(String::valueOf).collect(Collectors.joining(" ")));
    }

    static void operate(char[] bits, int prefixLength) {
        for (int i = 0; i < prefixLength; ++i) {
            bits[i] = (char) ('0' + '1' - bits[i]);
        }

        for (int i = 0, j = prefixLength - 1; i < j; ++i, --j) {
            char temp = bits[i];
            bits[i] = bits[j];
            bits[j] = temp;
        }
    }
}