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.math.*;
import java.util.*;


// @author : Dinosparton 

public class test {
	 
	   static class Pair{ 
		   long x;
		   long y;
		   
		   Pair(long x,long y){ 
			   this.x = x;
			   this.y = y;
			   
		   }
	   }
	  
	   static class Compare { 
		   
		     void compare(Pair arr[], int n) 
		    { 
		        // Comparator to sort the pair according to second element 
		        Arrays.sort(arr, new Comparator<Pair>() { 
		            @Override public int compare(Pair p1, Pair p2) 
		            { 
		                return (int)(p1.x - p2.x); 
		            } 
		        }); 
		  
//		        for (int i = 0; i < n; i++) { 
//		            System.out.print(arr[i].x + " " + arr[i].y + " "); 
//		        } 
//		        System.out.println(); 
		    } 
		} 
	  
	   static long gcd(long a, long b)
	    {
	      if (b == 0)
	        return a;
	      return gcd(b, a % b);
	    }
	   
	   static char [] reverse(char a[],int n){
			char b[]=new char[n];
			for(int i=0;i<n;i++) {
				a[i]=(a[i]=='0'?'1':'0');
				b[i]=a[i];
			}
			for(int i=0;i<n;i++)a[i]=b[n-i-1];
			return a;
		}
		
	   public static void main(String args[]) throws Exception { 
		
		   
		   Scanner sc = new Scanner(System.in);
		  
		   int tc = sc.nextInt();
		   
		   while(tc-->0) { 
			  
			   int n = sc.nextInt();
			   
			   char a[]=sc.next().toCharArray();
			   char b[]=sc.next().toCharArray();
			   
				ArrayList<Integer> list=new ArrayList<>();
				
				for(int i=n-1;i>=0;i--) {
					if(a[i]==b[i])continue;
					else {
						if(a[0]!=b[i]) {
							a=reverse(a,i+1);
							list.add(i+1);
						}
						else {
							a[0]=(a[0]=='0'?'1':'0');
							a=reverse(a,i+1);
							list.add(1);
							list.add(i+1);
						}
					}
				}
				
				System.out.print(list.size()+" ");
				for(int i:list)System.out.print(i+" ");
				System.out.println();
				
		   }
		  
	   }
}