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
from itertools import *
def count_inversion(sequence):
    return sum(sum(m<n for m in sequence[i+1:]) for i,n in enumerate(sequence))

n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

up = []
down = []
for i in range(n):
    if i % 2 == 0:
        up.append((a[i], i))
        down.append((b[i], i))
    else:
        up.append((b[i], i))
        down.append((a[i], i))

up.sort()
l = set([i for i in range(n)])

ans = 10000
for u in combinations(up, (n+1)//2):
    index = []
    value = []
    for val, ind in u:
        index.append(ind)
        value.append(val)

    d_index = l - set(index)
    d = []
    for i in d_index:
        d.append(down[i])
    d.sort()
    value.append(100)
    flag = 1
    for i in range(len(d)):
        if not value[i] <= d[i][0] <= value[i+1]:
            flag = 0
            break
    if flag == 0:
        continue

    for i in range(len(d)):
        index.insert(2*i+1, d[i][1])

    ans = min(ans, count_inversion(index))



if ans == 10000:
    print(-1)
else:
    print(ans)
