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<bits/stdc++.h>
#define ll long long
using namespace std;
int n,a[20],b[20],ans=1987654321;
struct st{int x;int k;};
st tem[20],tem2[20];
bool nondec()
{
    for(int i=0;i<n-1;i++)
    {
        if(tem[i].x>tem[i+1].x||tem[i].k+tem[i+1].k)
        {
            return false;
        }
    }
    return true;
}
void swfl(int x)
{
    st ret={tem[x].x,tem[x].k};
    tem[x]={tem[x+1].x,1-tem[x+1].k};
    tem[x+1]={ret.x,1-ret.k};
}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",a+i);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d",b+i);
    }
    for(int i=0;i<(1<<n);i++)
    {
        for(int j=0;j<n;j++)
        {
            if(i&(1<<j))
            {
                tem[j]={b[j],1};
                tem2[j]=tem[j];
            }
            else
            {
                tem[j]={a[j],0};
                tem2[j]=tem[j];
            }
        }
        int cnt=0;
        for(int j=0;j<n-1;j++)
        {
            if(tem[j+1].k)
            {
                swfl(j);
                cnt++;
            }
        }
        if(nondec())
        {
            ans=min(ans,cnt);
        }
        cnt=0;
        for(int j=0;j<n;j++)
        {
            tem[j]=tem2[j];
        }
        for(int j=n-1;j>=0;j--)
        {
            if(tem[j].k)
            {
                swfl(j-1);
                cnt++;
            }
        }
        if(nondec())
        {
            ans=min(ans,cnt);
        }
    }
    if(ans==1987654321)
    {
        printf("-1");
        return 0;
    }
    printf("%d",ans);
}