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:
//package CodeForces;

import java.util.ArrayList;
import java.util.Scanner;

public class PrefixFlip {
	
	static void reverse(char ch[],int k)
	{
		int i=0;int j=k;
		
		while(i<=j)
		{
			if(i==j)
			{
				if(ch[j]=='1')
				{
					ch[i]='0';
				}
				else
				{
					ch[i]='1';
				}
				return;
			}
			char temp=ch[j];
			if(ch[i]=='1')
			{
				ch[j]='0';
			}
			else
			{
				ch[j]='1';
			}
			if(temp=='0')
			{
				ch[i]='1';
			}
			else
			{
				ch[i]='0';
			}
			i++;j--;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t>0)
		{
			int n=s.nextInt();
			String str1=s.next();
			String str2=s.next();
			char ch1[]=str1.toCharArray();
			char ch2[]=str2.toCharArray();
			ArrayList<Integer> ans=new ArrayList<Integer>();
			int j=n-1;
			while(j>=0)
			{
				while(j>=0&&ch1[j]==ch2[j])
				{
					j--;
				}
				if(j<0)
				{
					break;
				}
				if(ch1[0]!=ch2[j])
				{
					ans.add(j+1);
					reverse(ch1,j);
					j--;
				}
				else
				{
					ans.add(1);
					if(ch1[0]=='0')
					{
						ch1[0]='1';
					}
					else
					{
						ch1[0]='0';
					}
				}
			}
			
			if(ans.size()==0)
			{
				System.out.println(2+" "+n+" "+n);
			}
			else
			{
				StringBuilder small=new StringBuilder();
				System.out.print(ans.size()+" ");
				for(int k=0;k<ans.size();k++)
				{
					if(k==ans.size()-1)
					{
						System.out.print(ans.get(k));
					}
					else
					{
						System.out.print(ans.get(k)+" ");
					}
				}
				System.out.println();
			}
			
//	       for(int k=0;k<ch1.length;k++)
//	       {
//	    	   System.out.print(ch1[k]);
//	       }
//	       System.out.println();
			t--;
		}

	}

}