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:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.util.*;
import java.io.*;
/**
 *
 * @author billy
 */
public class PrefixFlipHard {
    
    
    static class Pair implements Comparable<Pair>{
        int x;
        int y;

        Pair (int x0, int y0){
            x=x0;
            y=y0;

        }
        public int compareTo(Pair e){
            if (y>e.y){
                return 1;
            }
            else if (y<e.y){
                return -1;
            }
            else return 0;
        }
    }
     static String flip(String a, int ind){
        String temp = "";
        for (int i = ind-1; i>=0; i--){
            if (a.charAt(i)=='1'){
                temp+="0";
            }
            else{
                temp+="1";
            }
        }
        if (ind!=a.length() && ind>0){
            return temp + a.substring(ind);
        }
        
        else{
            return temp;
        }
    }
    static boolean check(String a, String b){
       for (int i =0; i<a.length(); i++){
           if (a.charAt(i) == b.charAt(i)){
               continue;
           }
           return false;
       }
       return true;
   }
        static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;


    public static void main(String[] args) throws IOException {
       
        for (int t = readInt(); t>0; t--){
            int n = readInt();
            String a = readLine();
            String b = readLine();
            ArrayList<Integer> ans = new ArrayList();
            
            int index = 0;
            for (int i = 0; i<n; i++){
                if(i%2==0){
                    if (a.charAt(index) != b.charAt(n-i-1)){
                        ans.add(n-i);
                        
                    }
                    else{
                        ans.add(1);//flip first
                        ans.add(n-i);//flips back
                    }
                }
                else{
                    if(a.charAt(n -index - 1) == b.charAt(n - i -1)){
                        ans.add(n-i);
                    }
                    else{
                        ans.add(1);
                        ans.add(n-i);
                    }
                    index++;
                }
            }
            System.out.print(ans.size() + " ");
            for (int i =0; i<ans.size(); i++){
                System.out.print(ans.get(i) + " ");
            }
            System.out.println();
        }
        
    }





    static String read () throws IOException {
        while (st == null || !st.hasMoreTokens())
           st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    //© Maillew DMOJ
    static long readLong () throws IOException {
        return Long.parseLong(read());
    }
    static int readInt () throws IOException {
        return Integer.parseInt(read());
    }
    static double readDouble () throws IOException {
        return Double.parseDouble(read());
    }
    static char readChar () throws IOException {
        return read().charAt(0);
    }
    static String readLine () throws IOException {
        return br.readLine().trim();
    }
}