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()))
INF = N**3
max_ab = max(max(A),max(B))+1
dp = [[INF] * max_ab for _ in range(2**N)]
dp[0][0]=0
for i in range(2**N):
  done = 0
  for j in range(N):
    if (i>>j)&1:
      done += 1
  
  for right_end in range(max_ab):
    if dp[i][right_end]==INF:
      continue
    for add_last in range(N):
      if (i>>add_last)&1:
        continue
      if abs(add_last-done)%2==0:
        add_last_surface = A[add_last]
      else:
        add_last_surface = B[add_last]
        
      if right_end>add_last_surface:
        continue
        
      num = 0
      for k in range(add_last):
        if (i>>k)&1==0:
          num+=1
          
      dp[i|1<<add_last][add_last_surface]=min(dp[i|1<<add_last][add_last_surface],
                                              dp[i][right_end]+num)
      
ans = min(dp[-1])
if ans==INF:
  print(-1)
else:
  print(ans)