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 PyPy3 (2.4.0) (judge verdict: Accepted)




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

INF = 1 << 20

# bitで１が立っているidxのカードがソートされて左詰めで並んでいるとき、その右端（＝最大のカード）がsとなる最小コスト
dp = [[INF for _ in range(51)] for _ in range(1 << N)]
dp[0][0] = 0

for bit in range(1 << N):

    # 既に並べてあるカードの枚数
    num_used = bin(bit).count("1")

    curr = num_used
    remaining = []
    for i in range(N):
        # bitのi番目に１が立っていない（＝まだ選ばれていない）とき
        if not(bit & (1 << i)):
            remaining.append((i, curr))
            curr += 1

    for s in range(51):
        # あるカードの並べ方について、右端をsにできないならスキップ
        if dp[bit][s] >= INF:
            continue

        for ini, curr in remaining:
            # iniの場所に最初いたカードを、num_usedの位置に持ってくるので、その偶奇によって表か裏かがわかる
            num = A[ini] if abs(num_used - ini) % 2 == 0 else B[ini]

            # numがsより小さい場合、整列したカードの最後尾につけられない
            if num >= s:
                # ある選び方で並んでいるカード群に、ini番目のカードを新たに最後尾に追加したとき、最後尾がnumとなる最小コスト
                dp[bit | (1 << ini)][num] = min(dp[bit | (1 << ini)][num], dp[bit][s] + abs(num_used - curr))

ans = INF
for i in range(51):
    ans = min(ans, dp[-1][i])

print(ans if ans < INF else -1)