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.*;
//import java.util.*;

/**
 * Created by Ayushi on 21 Jul 2020.
 * Problem:
 * Round:
 */

public class C {
    public static void main(String[] args) throws IOException {
        //StringTokenizer st;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine());
            StringBuilder a = new StringBuilder(br.readLine());
            //StringBuilder b = new StringBuilder(br.readLine());
            String b = br.readLine();
            StringBuilder c;
            String d;
            StringBuilder ans = new StringBuilder();
            int m = 0;

            for (int i = n-1; i >= 0; i--) {
                if (a.charAt(i) != b.charAt(i)) {
                    if (a.charAt(0) == b.charAt(i)) {
                        //System.out.print("1 ");
                        m++;
                        ans.append(1);
                        ans.append(" ");
                        if (a.charAt(0) == '0') a.replace(0, 1, "1");
                        else a.replace(0 ,1, "0");
                        //System.out.println(a.toString());
                    }

                    ans.append((i+1));
                    ans.append(" ");
                    m++;
                    c = new StringBuilder(a.substring(0, i+1));
                    //System.out.println(c.toString());
                    c.reverse();
                    d = c.toString();
                    //System.out.println(d);
                    d = d.replace('1', 'z');
                    d = d.replace('0', '1');
                    d = d.replace('z', '0');
                    //System.out.println(d);
                    a.replace(0, i+1, d);
                    //System.out.print(i + " ");
                    //System.out.println(a.toString());
                }
                /*else {
                    System.out.println("The digits at positions " + i + " are the same");
                    System.out.println("a.charAt(i) = " + a.charAt(i));
                    System.out.println("b.charAt(i) = " + b.charAt(i));
                }*/

            }
            //System.out.println(a.toString());
            System.out.println(m + " " + ans);
        }
        br.close();
    }
}