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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

INF = 1 << 20
dp = [[INF] * 51 for _ in range(1 << N)]
dp[0][0] = 0

for bit in range(1 << N):
    # 左詰めソートされている数
    con = bin(bit).count("1")

    # bitに含まれないカード
    ords = []
    iters = con
    for i in range(N):
        if not (bit & (1 << i)):
            ords.append([i, iters])
            iters += 1

    for s in range(51):
        if dp[bit][s] >= INF:
            continue

        for p in ords:
            # p[0]: 元々のindex, p[1]: dp[bit][s]の状態でのindex
            # con: bitの中身の数（既に詰まっているindexの最後尾）
            # |i-j|が偶数ならばA[i], 奇数ならばB[i]
            ns = A[p[0]] if abs(p[0] - con) % 2 == 0 else B[p[0]]

            # bitの中身がcon枚なので，bitに含まれないものはcon+1から順に詰める
            # 新しく詰める値nsが，既に詰まっている値の最大値sよりも大きい場合，
            # その値を後ろに詰める．
            if ns >= s:
                # 詰める時の最小の操作回数をdpで調べる．
                dp[bit|(1 << p[0])][ns] = min(dp[bit|(1 << p[0])][ns],
                                              dp[bit][s] + abs(con - p[1]))

ans = min(dp[-1][s] for s in range(51))
print(ans if ans < INF else -1)
