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.io.*;
import java.util.*;
	
public class C 
{	
	static int[][] mat,vis,cc;
	public static void main(String[] args)throws IOException 
	{	
		FastReader f=new FastReader();
		
		
		int test=f.nextInt();
		while(test-->0)
		{	
			StringBuffer sb = new StringBuffer();
			int n=f.nextInt();
			
			char c1[]=f.next().toCharArray();
			char c2[]=f.next().toCharArray();
			
			int count=0;
			for(int i=0;i<c1.length;i++)
			{	
				int idx=i+1;
				if(c1[i]!=c2[i])
				{
					sb.append(idx+" 1 "+idx+" ");
					count+=3;
				}
			}
			System.out.println(count+" "+sb);
			
		}
	}
	
	static class FastReader 
	{ 
	    BufferedReader br; 
	    StringTokenizer st; 

	    public FastReader() {
	    	br = new BufferedReader(new
	                 InputStreamReader(System.in)); 
	    }
	    String next() { 
	        while (st == null || !st.hasMoreElements()) { 
	            try{ 
	                st = new StringTokenizer(br.readLine()); 
	            } 
	            catch (IOException  e) { 
	                e.printStackTrace(); 
	            } 
	        } 
	        return st.nextToken(); 
	    } 
	    int nextInt() { 
	        return Integer.parseInt(next()); 
	    } 
	    long nextLong() { 
	        return Long.parseLong(next()); 
	    } 
	    double nextDouble() { 
	        return Double.parseDouble(next()); 
	    } 
	    String nextLine() { 
	        String str = ""; 
	        try{ 
	            str = br.readLine(); 
	        } 
	        catch (IOException e) { 
	            e.printStackTrace(); 
	        } 
	        return str; 
	    } 
	} 
}