Problem: 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

Solution (in PYTHON3):

Solution:
from itertools import combinations
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

even = [a[i] if i % 2 == 0 else b[i] for i in range(n)]
odd = [a[i] if i % 2 == 1 else b[i] for i in range(n)]
ans = 1001001001
for s in combinations(range(n), n // 2):
    # s ... dest index is odd
    s = set(s)
    t = set(range(n)) - s
    sdest = [(odd[i], i) for i in s]
    tdest = [(even[i], i) for i in t]
    sdest.sort()
    tdest.sort()
    origs = []
    z = []
    for i in range(n):
        if i & 1 == 0:
            origs.append(tdest[i//2][1])
            z.append(tdest[i//2][0])
        else:
            origs.append(sdest[(i-1)//2][1])
            z.append(sdest[(i-1)//2][0])
    if not all(x <= y for x, y in zip(z, z[1:])):
        continue
    dests = [0] * n
    for dest, orig in enumerate(origs):
        dests[orig] = dest
    cnt = 0
    for i in range(n):
        for j in range(i):
            cnt += dests[j] > dests[i]
    ans = min(cnt, ans)

if ans == 1001001001:
    ans = -1
print(ans)