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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main
{
    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;
        }
    }

    public static void main(String[] args)
    {
        FastReader sc=new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            String[] a = sc.nextLine().split("");
            String[] b = sc.nextLine().split("");
            int[] A = new int[n];
            int[] B = new int[n];
            for(int i=0;i<n;i++)
                A[i] = Integer.parseInt(a[i]);
            for(int i=0;i<n;i++)
                B[i] = Integer.parseInt(b[i]);
            boolean flag = true;
            for(int i=0;i<n;i++){
                if(A[i]!=B[i]){
                    flag = false;
                    break;
                }
            }
            if(flag)
                System.out.println(0);
            if(!flag) {
                ArrayList<Integer> ar = new ArrayList<>();
                if (A[0] != B[0]) {
                    if (A[0] == 1)
                        A[0] = 0;
                    else
                        A[0] = 1;
                    ar.add(1);
                }
                for (int i = 1; i < n; i++) {
                    if (A[i] != B[i]) {
                        A = inverse(A, i);
                        A = reverse(A, i + 1);
                        ar.add(i + 1);
                        A = inverse(A, 0);
                        ar.add(1);
                        A = inverse(A, i);
                        A = reverse(A, i + 1);
                        ar.add(i + 1);

                    }
                }
                System.out.print(ar.size() + " ");
                for (int i : ar)
                    System.out.print(i + " ");
                System.out.println();
            }
        }
    }
    static int[] reverse(int []a, int k)
    {
        for (int i = 0; i < k/2; i++)
        {
            int tempswap = a[i];
            a[i] = a[k - i - 1];
            a[k - i - 1] = tempswap;
        }
        return a;
    }
    static  int[] inverse(int[] a, int p){
        for(int i=0;i<=p;i++){
            if(a[i]==0)
                a[i]=1;
            else
                a[i]=0;
        }
        return a;
    }
}