Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
const ll mod=1000000007;
ll gcd(ll a,ll b) {return b ? gcd(b,a%b):a;}
ll lcm(ll c,ll d){return c/gcd(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
ll dp[(1<<18)+5][51];
int main(){
for(int i=0;i<=(1<<18);i++)for(int j=0;j<=50;j++)dp[i][j]=inf;
int n;
cin>>n;
V<int> a(n),b(n);
for(int i=0;i<n;i++)cin>>a[i];
for(int i=0;i<n;i++)cin>>b[i];
dp[0][0]=0;
for(int bit=0;bit<(1<<n);bit++){
  int cnt=0;
  for(int j=0;j<n;j++)if(bit&(1<<j))cnt++;
  V<P> d;
  int t=cnt;
   for(int j=0;j<n;j++)if(!(bit&(1<<j)))d.emplace_back(j,t++);
   for(int i=0;i<=50;i++){
     if(dp[bit][i]>=inf)continue;
     for(int j=0;j<d.size();j++){
       P now=d[j];
       int num;
       if(abs(now.fi-cnt)%2==0)num=a[now.fi];
       else num=b[now.fi];
       if(num>=i)chmin(dp[bit|(1<<now.fi)][num],dp[bit][i]+(now.se-cnt));
  }
  }
  }
  ll ans=inf;
  for(int i=0;i<=50;i++)chmin(ans,dp[(1<<n)-1][i]);
  if(ans>=inf)cout<<-1<<endl;
  else cout<<ans<<endl;
}