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 Python (3.4.3) (judge verdict: Runtime Error)

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;

signed main(){
  int n;cin >> n;
  vector<int> a(n);for(int i=0;i<n;i++) cin >> a[i];
  vector<int> b(n);for(int i=0;i<n;i++) cin >> b[i];
  int inf=10000000;
  int ans=inf;
  for(int i=0;i<(1<<n);i++){
    vector< vector<int> > x(n,vector<int>(3));
    for(int j=0;j<n;j++){
      if(((i>>j)&1)==1){
        x[j][0]=a[j];x[j][1]=j;x[j][2]=0;
      }else{
        x[j][0]=b[j];x[j][1]=j;x[j][2]=1;
      }
    }
    vector<int> y(n);
    for(int j=0;j<n;j++){y[j]=x[j][0];}
    sort(x.begin(),x.end());
    bool f=true;
    for(int j=0;j<n;j++){
      if(abs(x[j][1]-j)%2!=x[j][2]){f=false;break;}
    }
    //for(int j=0;j<n;j++){cout << x[j][0] << " " << x[j][1] << " " << x[j][2] << endl;}
    if(f){
      int c=0;
      for(int j=0;j<n;j++){
        for(int k=j;k>0;k--){
          if(y[k]>=y[k-1])break;
          else{swap(y[k],y[k-1]);c+=1;}
          if(c==ans)break;
        }
        if(c==ans)break;
      }
      //cout << ans <<endl;
      ans=min(ans,c);
    }
  }
  if(ans==inf){
    cout << -1 << endl;
  }else{
    cout << ans << endl;
  }
}