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.Arrays;
import java.util.Scanner;

public class Main {
	static final int N = 1100;

	static void op1(char a[]) {
		if (a[0] == '1')
			a[0] = '0';
		else
			a[0] = '1';

	}

	static void op2(int m, char a[]) {
		char tmp[] = Arrays.copyOf(a, a.length);
//		System.out.println(tmp);
		for (int i = m - 1; i >= 0; i--) {
			if (tmp[i] == '1')
				a[m - i - 1] = '0';
			else
				a[m - i - 1] = '1';
		}
//		System.out.println("---");
//		System.out.println(m);
//		System.out.println(tmp);
//		System.out.println(a);
//		System.out.println("***");
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while (t-- > 0) {
			ArrayList<Integer> ans = new ArrayList<Integer>();
			int n = sc.nextInt();
			sc.nextLine();
			char a[] = sc.nextLine().toCharArray();
			char b[] = sc.nextLine().toCharArray();
			for (int i = n - 1; i >= 0; i--) {
				if (a[i] == b[i])
					continue;
				if (a[0] == b[i]) {
					op1(a);
					ans.add(1);
					
				}
//				System.out.println(a);
				op2(i+1, a);
				ans.add(i + 1);
//				System.out.println(a);
			}
			System.out.print(ans.size());
			for (Integer x : ans) {
				System.out.print(" " + x);
			}
			System.out.println("");
		}

	}

}
/*

1
5
01011
11100

 */