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<stdio.h>
#include<string>
#include<utility>
#include<iostream>
#include<queue>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<bitset>

using namespace std;
typedef long long ll;
#define REP(i,n) for(int i=0;i<(int)(n);i++)

int mat[20][2]={};
int out[20]={};
int mot[20]={};
int rot[20]={};

int get(int num, int keta){
    REP(i,keta){
        int flag = num%2;
        out[i]=mat[i][flag];
        mot[i]=mat[i][flag];
        rot[i]=flag;
        num/=2;
    }
    sort(out,out+keta);
}

int main(){
    int n;
    int a,b;
    int visited[20]={};
    int rotate[20]={};
    int min = 100000;
    scanf("%d",&n);
    REP(i,n){
        scanf("%d",&a);
        mat[i][0]=a;
    }
    REP(i,n){
        scanf("%d",&b);
        mat[i][1]=b;
    }
    REP(i,2<<(n-1)){
        get(i,n);
        REP(j,20)visited[j]=0;
        REP(j,20)rotate[j]=0;
        int time=0;
        REP(j,n){
            int now=0;
            int num=out[j];
            int count=0;
            while(now<n){
                if(visited[now]==0 && mot[now]==num && rot[now]==(count%2+rotate[now])%2){
                    visited[now]=1;
                    time += count;
                    break;
                }
                else if(visited[now]==1){
                    now++;
                }
                else{
                    rotate[now] = 1-rotate[now];
                    now++;
                    count++;
                }
            }
        }
        int flag=1;
        REP(j,n)if(visited[j]==0)flag=0;
        if(flag==1)if(min>time)min=time;

        //REP(j,n)printf("%d,",out[j]);
        //if(flag)printf("   %d###\n",time);
        //else printf("   -1###\n");
    }
    if(min==100000)printf("-1\n");
    else printf("%d\n",min);
    return 0;
}