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 Div2_658_C1 {
	
	public static String rev(String s) {
		String ans ="";
		int l = s.length();
		for(int i = l-1;i>=0;i--) {
			char ch = s.charAt(i);
			if(ch=='0') {
				ans+="1";
			}
			else {
				ans+="0";
			}
		}
		return ans;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		for(int p = 0; p<t;p++) {
			int n = scan.nextInt();
			scan.nextLine();
			String s1 = scan.nextLine();
			String s2 = scan.nextLine();
			ArrayList<Integer> arr = new ArrayList<Integer>();
			for(int i = n-1;i>=0;i--) {
				char ch1 = s1.charAt(i);
				char ch2 = s2.charAt(i);
				if(ch1!=ch2) {
					char ch = s1.charAt(0);
					if(ch!=ch1) {
						arr.add(1);
						if(ch=='0') {
							s1 = "1"+s1.substring(1);
						}
						else {
							s1="0"+s1.substring(1);
						}
						arr.add(i+1);
						s1=rev(s1.substring(0,i+1));
					}
					else {
						arr.add(i+1);
						s1=rev(s1.substring(0,i+1));
					}
				}
			}
			int l = arr.size();
			System.out.print(l+" ");
			for(int i = 0;i<arr.size();i++) {
				System.out.print(arr.get(i)+" ");
			}
			System.out.println();
		}
	}

}