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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.concurrent.atomic.AtomicReference;

public class Main {


   public static void main(String[] args) {
//        Scanner sc = new Scanner(System.in);
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t=sc.nextInt();
        outer:while (t-->=1) {
            int n=sc.nextInt();
            String a=sc.next();
            String b=sc.next();
            ArrayList<Integer> ans= new ArrayList<>();
            for (int i=0;i<n;i++){
                if (a.charAt(i)!=b.charAt(i)){
                    ans.add(i+1);
                    ans.add(1);
                    ans.add(i+1);
                }
            }
            System.out.print(ans.size()+" ");
            for (int i=0;i<ans.size();i++){
                System.out.print(ans.get(i)+" ");
            }
            System.out.println();

            out.flush();
        }

        }
    static boolean find(int x){
        return (x!=0) && ((x&(x-1))==0);
    }



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

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long[] readArrayLong(long n) {
            long[] a=new long[(int)n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }


}