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: Wrong Answer)

#include<iostream>
#include<stdio.h>
#include<string>
#include<vector>
#include<queue>
#include <cmath>
#include<algorithm>
#include<stack>
#include<map>
#include<ctime>
#define rep(i,n) for (int i=0;i<n;i++)
//#include <bits/stdc++.h>
using namespace std;

int main(){
    int N,ans=10000;
    cin>>N;
    vector<vector<int> > A(2,vector<int>(N));
    rep(i,N) cin>>A[0][i];
    rep(i,N) cin>>A[1][i];
    for (long int i=0;i<(1<<N);i++){
        int n=0;
        vector<int> v(N);
        vector<int> f(N,0);
        for (int j=0;j<N;j++){
            if (i&(1<<j))  v[j]=A[0][j];
            else{
                v[j]=A[1][j];
                f[j]=1;
            }
        }
        bool flag=true,flag2=true;
        for (int j=0;j<N;j++){
            flag2=true;
            int m=0;
            for (int k=0;k<N;k++){
                if (k<j && v[j]<v[k]){
                    f[j]=f[j]+1;
                    n++;
                }
                if (k>j && v[j]>v[k]) f[j]=f[j]+1; 
                if (flag2 && k>j && v[k]==v[j]){
                    flag2=false;
                    m=k;
                } 
            }
            if (!flag2) f[m]=f[m]+f[j];
            if (flag2 && f[j]%2==1){
                flag=false;
                break;
            }
        }
        if (flag) ans=min(n,ans);
    }
    if (ans<10000) cout<<ans<<endl;
    else cout<<-1<<endl;
}