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>

using namespace std;
#define int long long
int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    int a[n];
    int b[n];
    for(int i=0;i<n;++i)
    {
        cin>>a[i];
    }
    for(int i=0;i<n;++i)
    {
        cin>>b[i];
    }
    int sz=(1<<n);
    int can[51][sz];
    for(int i=0;i<=50;++i)
    {
        can[i][0]=0;
    }
    for(int i=50;i>=0;i--)
    {
        for(int mask=1;mask<sz;++mask)
        {
            can[i][mask]=1e9;
            int s=0;
            for(int j=0;j<n;++j)
            {
                if(mask & (1<<j))
                {
                    ++s;
                }
            }
            int pos=(n-s);
            for(int j=0;j<n;++j)
            {
                if(mask & (1<<j))
                {
                    int t=abs(pos-j);
                    int curr;
                    if(t%2)
                    {
                        curr=b[j];
                    }
                    else
                    {
                        curr=a[j];
                    }
                    if(curr>=i)
                    can[i][mask]=min(t+can[curr][mask-(1<<j)],can[i][mask]);
                }
            }
        }
    }
    if(can[0][sz-1]!=(1e9))
    cout<<can[0][sz-1]/2;
    else
    cout<<(-1);
    return 0;
}