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.util.*;
public class file{
	public static void main(String[] args) {
		Scanner scn=new Scanner(System.in);
		int test=scn.nextInt();
		while(test-->0) {
			int n=scn.nextInt();
			scn.nextLine();
			String a=scn.nextLine();
			String b=scn.nextLine();
			if(a.equals(b)) {
				System.out.println("0");
			}else {
				StringBuilder sb=new StringBuilder();
				for(int i=n-1;i>=0;i--) {
					if(b.charAt(i)=='0')
						sb.append('1');
					else
						sb.append('0');
				}
				
				int c=0;
				String tem=sb.toString();
				StringBuilder ans=new StringBuilder();
				
				if(a.charAt(0)!=tem.charAt(0)) {
					c++;
					ans.append(1+" ");
				}
					
			int cou=0;
				for(int i=1;i<n;i++) {
					if(a.charAt(i) != tem.charAt(i)) {
						cou++;
						ans.append(i+1+" ");
						ans.append(1+" ");
						ans.append(i+1+" ");
					}
				}
				ans.append(n);
				c+=(cou*3)+1;
				System.out.println(c+" "+ans);
			}
		}
	}
}