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.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class PrefixFlip {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        char[] cb = in.next().toCharArray();
        char[] ca = in.next().toCharArray();
        int inv = 0;
        int lastInversePosition = n - 1;
        List<Integer> q = new ArrayList<>();
        for(int i = n - 1; i >= 0; i--){
            char c = '1';
            if (cb[lastInversePosition] == '1' && inv % 2 == 0){
                c = cb[lastInversePosition];
            }
            if (cb[lastInversePosition] == '0' && inv % 2 == 0){
                c = cb[lastInversePosition];
            }
            if (cb[lastInversePosition] == '0' && inv % 2 == 1){
                c = '1';
            }
            if (cb[lastInversePosition] == '1' && inv % 2 == 1){
                c = '0';
            }
            if (ca[i] == c){
                if (inv % 2 == 1){
                    lastInversePosition++;
                }else{
                    lastInversePosition--;
                }
            }else{
                if (inv % 2 == 0){
                    int first = lastInversePosition - i;
                    if (cb[first] == ca[i]){
                        q.add(1);
                    }
                    q.add(i + 1);
                    lastInversePosition = first + 1;
                }else{
                    int first = lastInversePosition + i;
                    if (cb[first] != ca[i]){
                        q.add(1);
                    }
                    q.add(i + 1);
                    lastInversePosition = first - 1;
                }
                inv++;
            }
        }
        out.print(q.size()+" ");
        for(int v : q){
            out.print(v+" ");
        }
        out.println();
    }
}