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 main():
  N = int(input())
  A = list(map(int,input().split()))
  B = list(map(int,input().split()))
  # M:カードの数字の最大値+1
  M = max(max(A),max(B))+1
  INF=float('inf')
  # dp[mask][a]:maskのカードが操作済で、そのうち右端のカードの数字がaであるときの操作回数の最小値
  dp = [[INF for _ in range(M)] for _ in range(1<<N)]
  dp[0][0] = 0
  
  for i in range(1<<N):
    # idx:操作済カードの枚数
    idx = 0
    for k in range(N):
      #k番目のカードが操作済みかどうか判定する。iでk番目にフラグがたってたら、操作済みとする。
      if (i>>k)&1 == 1:
        idx += 1
    for j in range(M):
      if dp[i][j] >= INF:
        continue
      for k in range(N):
        # k:右端の操作済カードの次に来るカード
        # kがすでに操作済の場合はcontinueする
        if (i>>k)&1:
          continue
        
        # kが何回操作されるかによってkの数字が表の数字になるか裏の数字になるかが決まる
        if abs(k-idx)%2 == 0:
          c = A[k]
        else:
          c = B[k]
  
        # kの数字広義単調増加にならない場合はcontinue
        if j > c:
          continue
        
        # t:kを右端の操作済カードの次に並べるための操作回数
        # それは、初期配置でkより左にあってまだ操作済でないカードの枚数と一致する
        t = 0
        for l in range(k):
          if (i>>l)&1 == 0:
            t += 1
        #print(i,k,idx,c,dp[i|(1<<k)][c],t)
        dp[i|(1<<k)][c] = min(dp[i|(1<<k)][c],dp[i][j]+t)
  
  ans = INF
  for i in range(M):
    ans = min(dp[(1<<N)-1][i],ans)
  print(ans if ans < INF else -1)

if __name__ == '__main__':
  main()