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.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author anmol
 */
public class fghef {
    public static void main(String []args){
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        outer:
        while(t-->0){
               //int n = s.nextInt();
               int n = s.nextInt();
               String s1 = s.next();
               String s2  =s.next();
               ArrayList list = new ArrayList();
               
               for(int i= s1.length()-1;i>=0;i--){
                   if(s1.charAt(i)==s2.charAt(i)) continue;
                   if(s1.charAt(0)==s1.charAt(i)) {
                       s1 = reverse(s1.substring(0,i+1)) + s1.substring(i+1,s1.length());
                       list.add(i+1);
                   }
                   else{
                       s1 = reverse(s1.substring(0,1)) + s1.substring(1,n);
                       list.add(1);
                       s1 = reverse(s1.substring(0,i+1)) + s1.substring(i+1,s1.length());
                       list.add(i+1);
                   }
               }
               System.out.print(list.size()+" ");
               for(int i=0;i<list.size();i++){
                   System.out.print(list.get(i)+ " ");
               }
               System.out.println();

        }
    }
    public static String reverse(String str){
        char ch[] = str.toCharArray();
        for(int i=0;i<ch.length;i++){
            if(ch[i]=='0') ch[i]='1';
            else ch[i]='0';
        }
        for(int i=0;i<ch.length/2;i++){
            char temp = ch[i];
            ch[i] = ch[ch.length-i-1];
            ch[ch.length-i-1]  = temp;
        }
        String ans = String.valueOf(ch);
        return ans;
    }
    


    
}