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 Demo111 {
	
	
	
	
	
	public static void main(String [] args)
	{
		Scanner in  = new Scanner(System.in);
		int t =in.nextInt();
		
		for(int i =0;i<t;i++)
		{
			int n =in.nextInt();
			
			String a =in.next();
			String b =in.next();
			char [] aAr =a.toCharArray();
			char [] bAr  =b.toCharArray();
			char []cAr  =new char [n];
			
			for(int j =0;j<n;j++)
			{
				cAr[j] =aAr[j];
				
			}
			
			long steps =0;
			ArrayList<Integer> path =new ArrayList<>();
			
			for(int j =n-1;j>-1;j--)
			{
				if(aAr[j]!=bAr[j])
				{
					if(aAr[0]!=bAr[j])
					{
						steps++;
						path.add(j+1);
						for(int k =j;k>-1;k--)
						{
							if(cAr[j-k]=='0')
							{
								aAr[k] ='1';
							}
							else {
								aAr[k] ='0';
							}
							
						}
						
					}
					
					else {
						steps =steps+2;
						path.add(1);
						path.add(j+1);
						
						if(aAr[0]=='0')
						{
							aAr[0] ='1';
						}
						else {
							aAr[0] ='0';
						}

						for(int k =j;k>-1;k--)
						{
							
							if(cAr[j-k]=='0')
							{
								aAr[k] ='1';
							}
							else {
								aAr[k] ='0';
							}
							
						}
						
						
						
					}
					
					
				}
				
				for(int k =0;k<n;k++)
				{
					cAr[k] =aAr[k];
				}
				
				
				
			}
			
			System.out.print(steps+" ");
			
			for(Integer single:path)
			{
				System.out.print(single+" ");
			}
			System.out.println();
			
			
			
			
			
		}
		

		
	
            }
	
}