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:
# -*- coding: utf-8 -*-

import sys
from itertools import accumulate

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7

def pop_count(i):
    i = i - ((i >> 1) & 0x55555555)
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
    i = (i + (i >> 4)) & 0x0f0f0f0f
    i = i + (i >> 8)
    i = i + (i >> 16)
    return i & 0x3f

N = INT()
cards = [[]] * 2
cards[0] = LIST()
cards[1] = LIST()

# dp[S][i] := 既に使用したカードの集合がSで、最後に使った(現在の右端の)カードがiの時の最小合計転倒数
dp = list2d(1<<N, N, INF)
for i in range(N):
    # 一番左は何が来ても転倒数0
    dp[1<<i][i] = 0
for S in range(1, 1<<N):
    popcnt = pop_count(S)
    # 各位置より元indexが右側にある出現済要素の個数(転倒数)を前計算しておく
    acc = [0] * N
    for i in range(N):
        if S & 1<<i:
            acc[i] = 1
    # 後ろから累積和
    acc = list(accumulate(acc[::-1]))[::-1]
    for i in range(N):
        if dp[S][i] == INF or not S & 1<<i:
            continue
        # 遷移元の右端カードの表裏を確定
        cur = cards[(i&1)^(1-(popcnt&1))][i]
        for j in range(N):
            if S & 1<<j:
                continue
            # 遷移先の右端カードの表裏を確定
            nxt = cards[(j&1)^(popcnt&1)][j]
            # 昇順が崩れないなら遷移させてOK
            if cur <= nxt:
                # 転倒数を追加して遷移
                dp[S|1<<j][j] = min(dp[S|1<<j][j], dp[S][i] + acc[j])
ans = INF
for i in range(N):
    ans = min(ans, dp[(1<<N)-1][i])
if ans == INF:
    print(-1)
else:
    print(ans)