#!/usr/bin/env python3

import sys
from typing import Any, Callable, Deque, Dict, List, Mapping, Optional, Sequence, Set, Tuple, TypeVar, Union
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)



def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79e+308
    # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def isp():   return input().split()
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())


    
    n = ii()
    A, B = lmi(), lmi()
    L = [A, B]
    M = max(max(A), max(B))

    dp = [[inf] * (M+1) for _ in range(2 ** n)]
    dp[0][0] = 0

    for S in range(2 ** n):
        for k in range(M + 1):
            if dp[S][k] != inf:
                cnt = 0    # 何枚のカードがすでにあるか
                for i in range(n):
                    if S & (1 << i):
                        cnt += 1
                for i in range(n):
                    if not (S & (1 << i)):
                        card_front = L[(i - cnt) % 2][i]
                        if card_front < k:
                            continue
                        inv = 0
                        for j in range(i + 1, n):
                            if S & (1 << j):
                                inv += 1
                        dp[S | (1 << i)][card_front] = min(dp[S + (1 << i)][card_front], dp[S][k] + inv)

    # for line in dp:
    #     print(*line)

    ans = min(dp[S])
    if ans == inf:
        print(-1)
    else:
        print(ans)




if __name__ == "__main__":
    main()



"""
想定解は bit DP.

勉強のため解説のような bit DP で実装.
以下方針

- 並べ方を決めれば操作によらずどの数字が表に並ぶか決定可能
- 並べ方を全探索してそれぞれ昇順判定、転倒数計算をすると O(N! * N * NlgN)
- dp[S][k] = (すでに並べた集合 S, 最後のカードインデックス k のときのここまでの転倒数) とする
    - dp[0][0] = 0
    - それ以外 inf
    - 配る DP. dp[S][k] (k in S) から配布する. S に含まれぬカードインデックスについてそれを末尾に並べられるか判定
      並べられる場合、そのカードの転倒数を計算。dp[S+i][i] をもとの値との最小値で更新
    - 基本ループは 2^N, N の二重ループ。遷移のところでさらに O(N) の走査が入る。トータルで O(2^N * N^2)
"""

Measured memory_bytes: 289024