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 C1 {
    public static int[] perform(int[] temp, int length) {
        int[] result = new int[temp.length];
        for(int i = 0; i <= length; i++) {
            if(temp[i] == 0) {
                result[i] = 1;
            } else result[i] = 0;
        }
        for(int i = length + 1; i < temp.length; i++) {
            result[i] = temp[i];
        }
        int[] answer = new int[temp.length];
        for(int i = 0; i <= length; i++) {
            answer[i] = result[length - i];
        }
        for(int i = length + 1; i < temp.length; i++) {
            answer[i] = result[i];
        }
        return answer;
    }
    public static void printArray(int[] temp){
        for(Integer i : temp) {
            System.out.print(i + " ");
        }
        System.out.println();
    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        for(int i = 0; i < t; i++) {
            int n = input.nextInt();
            input.nextLine();
            String temp1 = input.nextLine();
            String temp2 = input.nextLine();
            int[] one = new int[n];
            int[] two = new int[n];
            for(int k = 0; k < n; k++) {
                one[k] = Integer.parseInt(temp1.substring(k, k + 1));
            }
            for(int k = 0; k < n; k++) {
                two[k] = Integer.parseInt(temp2.substring(k, k + 1));
            }
            int length = 0;
            ArrayList<Integer> operations = new ArrayList<>();
            for(int k = n - 1; k >= 0; k--) {
                if(one[k] != two[k]) {
                    if(one[0] == two[k]) {
                        length++;
                        operations.add(1);
                        if(one[0] == 0) {
                            one[0] = 1;
                        } else one[0] = 0;
                    }
                    one = perform(one, k);
                    length++;
                    operations.add(k + 1);
                }
            }
            System.out.print(length);
            for(Integer k : operations) {
                System.out.print(" " + k);
            }
            System.out.println();
        }
    }
}