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.
//package round_658_div2;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class C2 {

	public static void main (String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());
		
		while(T > 0) {
			int N = Integer.parseInt(br.readLine());
			String str1 = br.readLine();
			String str2 = br.readLine();
			
			solve(str1, str2);
			
			T--;
		}
	}
	
	
	private static void solve(String str1 , String str2) {
		
		List<Integer> list1 = new ArrayList<>();
		List<Integer> list2 = new ArrayList<>();
		
		str1 += "0";
		str2 += "0";
		
		for(int i = 0 ; i < str1.length()-1 ; i++) {
			if(str1.charAt(i) != str1.charAt(i+1)) list1.add(i+1);
			if(str2.charAt(i) != str2.charAt(i+1)) list2.add(i+1);
		}
		
		Collections.reverse(list2);
		
		System.out.print((list1.size()+list2.size())+ " ");
		
		for(int a : list1) {
			System.out.print(a+" ");
		}
		
		for(int a : list2) {
			System.out.print(a+" ");
		}
		
		System.out.println();
	}
	
	private static  boolean allOne(String str) {
		for(char c : str.toCharArray()) {
			if(c != '1')
				return false;
		}
		
		return true;
	}
	
	private static String process(String str, int pre) {
		char[] chars = str.toCharArray();
		
		for(int i = 0; i < pre ; i++) {
			if(chars[i] == '0') {
				chars[i] = '1';
			}else {
				chars[i] = '0';
			}
		}
		str = new String(chars);
		
		StringBuffer sb = new StringBuffer(str.substring(0, pre));
		
		sb.reverse();
		
		str = sb.toString() + str.substring(pre);
		
		return str;
	}
}
