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
# reference -> https://atcoder.jp/contests/keyence2020/submissions/9649881
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

INF = 2**19

# dp[bit][s]
#   ビットに1が立っているカードはソートされて左から並んでいる
#   そのとき最後の値をsにするのに必要な最小操作回数
# 1 <= A,B <= 50, bitの立ち方が 0...0 ~ 1...1 の2**N
dp = [[INF for _ in range(50 + 1)] for _ in range((1<<N))]
# 0枚選んでいて右端が0は何もしてないのでコスト0
dp[0][0] = 0

for bit in range(1<<N):
  num_used = bin(bit).count("1") # 既に並んでいるカードの枚数
  idx = num_used
  remaining = [] # まだ並べていないカードがどれか
  for i in range(N):
    if not (bit & (1 << i)):
      # (初期位置, 現在位置)
      #   - 初期位置 : 最初の状態で左から何番目にあるか
      #   - 現在位置 : bitで1が立っているカードを並べたときに左から何番目にあるか
      remaining.append((i, idx))
      idx += 1
  for s in range(51):
    # bitの集合かつ、その右端にsの状態が作れないならスキップ
    if dp[bit][s] >= INF:
        continue
    for init_pos, current_pos in remaining:
      # 最後尾まで移動させる際の交換回数に応じてA, Bの適切な方を選ぶ
      number = [A[init_pos], B[init_pos]][abs(init_pos-num_used)%2]
      if number >= s: # 現段階の最後尾の値以上であればOK
        dp[bit | (1<<init_pos)][number] = min(
          dp[bit | (1<<init_pos)][number], dp[bit][s] + abs(num_used - current_pos)
        )


# 全てのbitに1が立っている（全て並べ切った）時の最小コストが答え
ans = min([dp[-1][s] for s in range(51)])

print(ans if ans < INF else -1)