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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
typedef pair<int,int> Pi;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define fi first
#define se second
#define endl "\n"

template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}
template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}
template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){
	int len=d.size();
	rep(i,len){
		s<<d[i];if(i<len-1) s<<" ";
	}
	return s;
}
template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){
	int len=d.size();
	rep(i,len){
		s<<d[i]<<endl;
	}
	return s;
}
template<typename T> ostream& operator<<(ostream& s,const set<T>& v){
	s<<"{ ";
	for(auto itr=v.begin();itr!=v.end();++itr) {
		if (itr!=v.begin()) {s<< ", ";}
		s<<(*itr);
	}
	s<<" }";
	return s;
}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){
	s<<"{"<<endl;
	for(auto itr=m.begin();itr!=m.end();++itr){
		s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;
	}
	s<<"}"<<endl;
	return s;
}

const ll mod=1e9+7;
const ll inf=1e17;
const int INF=1e9;
const double PI=acos(-1);
const double EPS=1e-10;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	vector<int> a(n),b(n);
	rep(i,n) cin>>a[i];
	rep(i,n) cin>>b[i];
	vector<vector<int>> dp((1<<n)+1,vector<int>(55,INF));
	dp[0][0]=0;
	rep(i,1<<n){
		int cnt=__builtin_popcount(i);//並びが決まっているカードの個数
		int now=cnt;
		vector<Pi> ret;
		rep(j,n){
			if(!(i & (1<<j))){//並びがまだ決まってないカード
				ret.push_back(Pi(j,now++));//それが現在左から何番目にあるか
			}
		}
		rep(j,55){
			if(dp[i][j]>=INF) continue;
			for(auto itr:ret){
				int ne;
				if(abs(itr.fi-cnt)%2==0) ne=a[itr.fi];
				else ne=b[itr.fi];
				if(ne>=j){
					chmin(dp[i|(1<<itr.fi)][ne],dp[i][j]+abs(itr.se-cnt));
				}
			}
		}
	}
	int ans=INF;
	rep(j,55){
		chmin(ans,dp[(1<<n)-1][j]);
	}
	if(ans<INF) cout<<ans<<endl;
	else cout<<-1<<endl;
}