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()))

res = []
# 0 = A , 1 = B
for bit in range(2 ** n):
    if str(bin(bit)).count('1') % 2 == 1:
        continue

    odds = []
    evens = []
    for i in range(n):
        if (bit >> i) & 1:
            card = B[i]
            check = (i + 1) % 2

        else:
            card = A[i]
            check = i % 2

        if check == 1:
            odds.append([card, i+1])
        else:
            evens.append([card, i+1])

    if 0 <= len(evens) - len(odds) <= 1:
        odds.sort(reverse=True)
        evens.sort(reverse=True)
        now = 0
        pattern = []
        for _ in range(len(evens)):
            num1, idx1 = evens.pop()

            if now <= num1:
                now = num1
                pattern.append(idx1)
                if odds:
                    num2, idx2 = odds.pop()
                    if now <= num2:
                        now = num2
                        pattern.append(idx2)
                    else:
                        break

            else:
                break
        else:
            res.append(pattern)


if not res:
    print(-1)
    exit()

ans = []
for pattern in res:
    count = 0
    for i in range(n-1):
        num = pattern[i]
        check = pattern[i+1:]
        for t in check:
            if t < num:
                count += 1

    ans.append(count)

print(min(ans))