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.*;
public class c {
	static void convert(char[] a , int l)
	{
		if (l==0) 
		{
			a[0] = a[0]=='0'?'1':'0';
			return;
		}
		char c = '0';
		for (int i=0,j=l;i<=j ; i++ ,j--)
		{
			a[i] = a[i]=='0'?'1':'0';
			if (i!=j)
			a[j] = a[j]=='0'?'1':'0';
			c = a[i];
			a[i] = a[j];
			a[j] = c;
		}
	}
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		while (t-->0)
		{
			int n = s.nextInt();
			char[] a1 = s.next().toCharArray();
			char[] a2 = s.next().toCharArray();
			StringBuilder res = new StringBuilder();
			int count = 0;
			for (int i = a1.length-1 ; i>=0;)
			{
				if (a1[i]==a2[i]) {i--; continue;}
				else {
					if (a1[0]==a2[i]) {
						convert(a1 , 0);
						res.append(1+" ");
						count++;
					}
					convert(a1 , i);
					res.append((i+1)+" ");
					count++;
				}
			}
			System.out.print(count+" ");
			System.out.println(res);
		}
	}
}