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:
def tentousu(a):
  ans=0
  for i in range(len(a)):
    for j in range(i+1,len(a)):
      if a[i]>a[j]:ans+=1
  return ans
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
for i in range(1,n,2):a[i],b[i]=b[i],a[i]
ans=inf=float('inf')
for i in range(2**n):
  s=bin(i)[2:].zfill(n)
  if s.count("0")!=n//2:continue
  ac=[]
  bc=[]
  for j in range(n):
    if s[j]=="1":ac.append((a[j],j))
    else:bc.append((b[j],j))
  ac.sort()
  bc.sort()
  x=[]
  for j in range(n):
    if j%2:x.append(bc[j//2])
    else:x.append(ac[j//2])
  f=False
  for j in range(1,n):
    if x[j-1][0]>x[j][0]:f=True
  if f:continue
  m=[]
  for j in range(n):m.append(x[j][1])
  ans=min(ans,tentousu(m))
if ans==inf:ans=-1
print(ans)