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: Time Limit Exceeded)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> l_l;
typedef pair<int,int> i_i;
const ll mod=1e9+7;
const int inf=1<<30;

const int MAX_N=100010;
int n,bit[MAX_N+1];

int sum(int i){
  int s=0;
  while(i>0){
    s+=bit[i];
    i-=(i&-i);
  }
  return s;
}

void add(int i,int x){
  while(i<=n){
    bit[i]+=x;
    i+=(i&-i);
  }
}

ll invnum(vector<int> a){
  rep(j,n+1) bit[j]=0;
  ll ans=0;
  for(int j=0;j<n;j++){
    ans+=j-sum(a[j]);
    add(a[j],1);
  }
  return ans;
}

int main(){
  cin >> n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  vector<int> b(n);
  rep(i,n) cin >> b[i];
  vector<int> v(n);
  rep(i,n) v[i]=i+1;
  ll ans=inf;
  do{
    vector<int> c(n);
    rep(i,n){
      if((abs(i+1-v[i]))%2==0) c[i]=a[v[i]-1];
      else c[i]=b[v[i]-1];
    }
    bool f=1;
    rep(i,n-1){
      if(c[i]>c[i+1]) f=0;
    }
    if(f) chmin(ans,invnum(v));
  }while(next_permutation(all(v)));
  if(ans==inf) ans=-1;
  cout << ans << endl;
}