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
class BinaryIndexedTree():  #1-indexed
    def __init__(self, n):
        self.n = n
        self.tree = [0 for _ in range(n + 1)]

    def sum(self, index):
        res = 0
        while index:
            res += self.tree[index]
            index -= index & -index
        return res

    def add(self, index, x):
        while index <= self.n:
            self.tree[index] += x
            index += index & -index

    def bisect_left(self, x):
        if x <= 0: return 0
        res, tmp = 0, 2**((self.n).bit_length() - 1)
        while tmp:
            if res + tmp <= self.n and self.tree[res + tmp] < x:
                x -= self.tree[res + tmp]
                res += tmp
            tmp >>= 1
        return res + 1


INF = 10**18

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

res = INF

for i in range(2**N):
    front = []
    back = []
    for j in range(N):
        if (i >> j) & 1:
            front.append((A[j], j))
        else:
            back.append((B[j], j))
    odd = []
    even = []
    for x, j in front:
        if j % 2 == 0:
            odd.append((x, j))
        else:
            even.append((x, j))
    for x, j in back:
        if j % 2 != 0:
            odd.append((x, j))
        else:
            even.append((x, j))
    if len(even) != N // 2:
        continue
    odd.sort()
    even.sort()
    seq = []
    for j in range(N):
        if j % 2 == 0:
            seq.append(odd[j // 2])
        else:
            seq.append(even[j // 2])
    for j in range(N - 1):
        if seq[j][0] > seq[j + 1][0]:
            break
    else:
        bit = BinaryIndexedTree(N)
        count = 0
        for j in range(N):
            bit.add(seq[j][1] + 1, 1)
            count += j + 1 - bit.sum(seq[j][1] + 1)
        res = min(res, count)

print(res if res != INF else -1)