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.util.*;
public class c1382 {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int tests = in.nextInt();
        for (int i = 0 ; i < tests; i++){
            in.nextInt();
            in.nextLine();
            String a = in.next();
            String b = in.next();
            List<Integer> ret = eval(a, b);
            for (int j : ret){
                System.out.print(j + " ");
            }
            System.out.println();
        }
        in.close();
    }
    private static List<Integer> eval(String a, String b){
        List<Integer> ret = new ArrayList<>();
        ret.add(0);
        StringBuilder sA = new StringBuilder(a);
        StringBuilder sB = new StringBuilder(b);
        if (a.equals(b)) {
            return ret;
        }
        boolean lense = true;
        for (int i = a.length() - 1; i >= 0; i--){
            char first = flip(sA.charAt(0), lense);
            char at = flip(sA.charAt(i), lense);
            if (at == sB.charAt(i)) continue;
            if (first == sB.charAt(i)){
                sA.replace(0, 1, "" + flip(first, false));
                ret.set(0, ret.get(0) +1);
                ret.add(1);
            }
            lense = !lense;
            StringBuilder reversed = new StringBuilder(sA.substring(0, i+1));
            reversed.reverse();
            sA.replace(0, i+1, reversed.toString());
            ret.set(0, ret.get(0) + 1);
            ret.add(i + 1);
        }
        return ret;
    }
    private static char flip(char c, boolean lense){
        if (lense) return c;
        return c == '1' ? '0' : '1';
    }
}