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()))
ans = 10**10
for bit in range(1<<n):
  l = a[:]
  rev = [0]*n
  for i in range(n):
    if bit & 1<<i:
      rev[i] = 1
      l[i] = b[i]
  if sum(rev) % 2:
    continue
  odd = list()
  even = list()
  for i in range(n):
    if rev[i]^(i&1):
      even.append((l[i],i))
    else:
      odd.append((l[i],i))
  if len(even) != n//2:
    continue
  odd.sort()
  even.sort()
  l = list()
  for x,y in zip(odd,even):
    l.append(x)
    l.append(y)
  if n&1: l.append(odd[-1])
  fl = 0
  for x,y in zip(l,l[1:]):
    if x[0] > y[0]:
      fl = 1
      break
  if fl: continue
  f = [x[1] for x in l]
  # print('ok', bit, even, odd)
  cnt = 0
  g = [0]*n
  for x in f:
    cnt += sum(g[x:])
    g[x] = 1
  if ans > cnt:
    ans = cnt
if ans > 10**9:
  ans = -1
print(ans)