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:
import itertools

n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = [i for i in range(n)]

def add(k, x):
  while k <= n:
    data[k] += x
    k += k & -k
def get(k):
  s = 0
  while k:
    s += data[k]
    k -= k & -k
  return s
ans = 10 ** 18
for i in itertools.combinations(c,n//2):
  j = list(set(c)-set(i))
  even = sorted([(a[k],k+1) if k % 2 else (b[k],k+1) for k in i])
  odd = sorted([(b[k],k+1) if k % 2 else (a[k],k+1) for k in j])
  prev = -1
  A = []
  for x in range(n):
    Y,X = divmod(x,2)
    if X % 2:
      now0,now1 = even[Y][0],even[Y][1]
    else:
      now0,now1 = odd[Y][0],odd[Y][1]
    if prev > now0:
      break
    else:
      A.append(now1)
    prev = now0
  else:
    data = [0]*(n+1)
    tmp = 0
    for i,z in enumerate(A):
      tmp += (n-1-i) - get(z)
      add(z,1)
    ans = min(ans,tmp)
if ans == 10 ** 18:
  print(-1)
else:
  print(ans)