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.
import java.util.*;
import java.util.Scanner;
import java.io.*;
import javax.lang.model.util.ElementScanner6; 
import static java.lang.System.out;


public class A11381
{

    static int arr[];
    
    public static void main(String args[])
    {  

        FastReader in=new FastReader();
        PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int tc=1;
     
        tc=in.nextInt();
        while(tc-->0)
        {
            
            int n=in.nextInt();

            String S=in.nextLine();
            String T=in.nextLine();

            int arr[]=new int[n];

            for(int i=0;i<n;i++)    if(S.charAt(i)!=T.charAt(i))    arr[i]=1;

       
            int count=0;

            ArrayList<Integer> al=new ArrayList<>();

            for(int i=0;i<n;i++)              
            {
                if(arr[i]==1)
                {
                    count+=3;
                    al.add(i+1);
                    al.add(1);
                    al.add(i+1);
               
                }
            }

            pr.print(count+" ");
            for(int i : al)pr.print(i+" ");
            pr.println();



        }
        pr.flush();
    }


    

    static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
    
    
	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
    }

    
    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; 
        } 
    } 
}