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()))
allpat=[]
todo=[[]]
while todo:
  pat=todo.pop()
  if len(pat)==n//2:
    allpat.append(pat)
    continue
  s=0
  if pat:s=max(pat)+1
  for i in range(s,n):
    todo.append(pat+[i])

def func(pat):
  # カードを最終的に偶奇どちらのidxになるかで分類
  # pat内のidxが最後に奇数のidxになるとする
  ablev=[]
  ablod=[]
  for i in range(n):
    if i in pat:
      if i%2==0:
        ablod.append([i,b[i]])
      else:
        ablod.append([i,a[i]])
    else:
      if i%2==0:
        ablev.append([i,a[i]])
      else:
        ablev.append([i,b[i]])
  ablev.sort(key=lambda x:x[1]*20+x[0])
  ablod.sort(key=lambda x:x[1]*20+x[0])
  now=0
  ary=[]
  for i in range(n//2):
    if ablev[i][1]>=now:
      now=ablev[i][1]
      ary.append(ablev[i][0])
    else:
      return -1
    if ablod[i][1]>=now:
      now=ablod[i][1]
      ary.append(ablod[i][0])
    else:
      return -1
  if n%2==1:
    if ablev[n//2][1]>=now:
      now=ablev[i][1]
      ary.append(ablev[n//2][0])
    else:
      return -1
  ret=0
  for i in range(n-1):
    for j in range(i+1,n):
      if ary[i]>ary[j]:ret+=1
  return ret

ans=float('inf')
for pat in allpat:
  t=func(pat)
  if t>=0:
    ans=min(ans,t)
print(ans if ans!=float('inf') else -1)
