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:
// This template code suggested by KT BYTE Computer Science Academy
//   for use in reading and writing files for USACO problems.

// https://content.ktbyte.com/problem.java

import java.util.*;
import java.io.*;

public class PrefixFlipHard {

    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int t = in.nextInt();

        for (int i = 0; i < t; i++) {
            int n = in.nextInt();

            char[] bin1 = in.next().toCharArray();
            char[] bin2 = in.next().toCharArray();

            ArrayDeque<Integer> deque = new ArrayDeque<>();

            for (int j = 0; j < n; j++) {
                if (bin1[j] == '1') deque.add(1);
                else deque.add(0);
            }

            int[] destination = new int[n];

            for (int j = 0; j < n; j++) {
                if (bin2[j] == '1') destination[j] = 1;
                else destination[j] = 0;
            }

            boolean reversed = false;

            ArrayList<Integer> steps = new ArrayList<>();

            for (int j = n-1; j >= 1; j--) {
                if (reversed && deque.peekFirst() != destination[j]) {
                    deque.removeFirst();
                }

                else if (!reversed && deque.peekLast() == destination[j]) {
                    deque.removeLast();
                }

                else {
                    if (deque.peekFirst() != deque.peekLast()) {
                        // need to reverse the "first" value

                        if (reversed) {
                            int val = deque.removeLast();
                            deque.addLast((val + 1) % 2);
                        }

                        else {
                            int val = deque.removeFirst();
                            deque.addFirst((val + 1) % 2);
                        }

                        steps.add(1);
                    }

                    steps.add(j+1);

                    reversed = !reversed;

                    if (reversed) {
                        deque.removeFirst();
                    }
                    else {
                        deque.removeLast();
                    }
                }
            }

            if (reversed && deque.peekFirst() == destination[0]) {
                steps.add(1);
            }

            else if (!reversed && deque.peekFirst() != destination[0]){
                steps.add(1);
            }

            out.print(steps.size() + " ");

            for (int j = 0; j < steps.size(); j++) {
                out.print(steps.get(j));

                if (j != steps.size() - 1) {
                    out.print(" ");
                }
            }

            out.println();
        }

        out.close();
    }
}