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()))
AB = (A, B)
INF = 10**18

ans = INF
for state in range(1 << N):
    even = []
    odd = []

    for i in range(N):
        isReverse = (state & (1 << i)) >> i
        val = AB[isReverse][i]

        if i % 2 == 0:
            (even, odd)[isReverse].append((val, i))
        else:
            (odd, even)[isReverse].append((val, i))

    if len(even) != -(-N // 2) or len(odd) != N // 2:
        continue

    even.sort()
    odd.sort()
    odd.append((-1, -1))
    V = []
    P = []
    for (l, i), (r, j) in zip(even, odd):
        V.append(l)
        P.append(i)
        if r >= 0:
            V.append(r)
            P.append(j)

    if not all(l <= r for l, r in zip(V, V[1:])):
        continue

    cnt = 0
    for i, p in enumerate(P):
        for q in P[:i]:
            if p < q:
                cnt += 1
    ans = min(ans, cnt)

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