Problem:
Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

def solve(not_used, index, cur_elem, cost, cur_min):
    #print("\t" * index, cur_elem, cost)
    if index == N:
        return cost
    for i in not_used:
        if (index - i) % 2 == 0:
            if A[i] >= cur_elem:
                new_cost = cost + abs(i - index)
                if new_cost < cur_min:
                    cur_min = min(cur_min, solve(not_used - {i}, index + 1, A[i], new_cost, cur_min))
        else:
            if B[i] >= cur_elem:
                new_cost = cost + abs(i - index)
                if new_cost < cur_min:
                    cur_min = min(cur_min, solve(not_used - {i}, index + 1, B[i], new_cost, cur_min))
    return cur_min

ans = solve({i for i in range(N)}, 0, -1, 0, 100000)
if ans == 100000:
    print(-1)
else:
    print(ans // 2)