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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=5e5;
int x[20],rev[100],a[20][2],f[N],n;
int calc(int state)
{
	for(int i=1;i<=n;i++)x[i]=a[i][state>>(i-1)&1];
	for(int i=1;i<=n;i++)rev[x[i]]=0;
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		int t=state>>(i-1)&1;
		for(int j=1;j<i;j++)t^=x[j]>x[i];
		for(int j=i+1;j<=n;j++)
			if(x[j]<x[i])++ans,t^=1;
		rev[x[i]]=rev[x[i]]<<1|t;
	}
	for(int i=1;i<=n;i++)ans+=f[rev[x[i]]],rev[x[i]]=0;
	return ans;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i][0]);
	for(int i=1;i<=n;i++)scanf("%d",&a[i][1]);
	f[0]=0,f[1]=100000;
	for(int S=2;S<(1<<n);S++)
		f[S]=(S&1)?f[S>>1^1]+1:f[S>>1];
	int ans=100000;
	for(int S=0;S<(1<<n);S++)ans=min(ans,calc(S));
	if(ans==100000)cout<<"-1"<<endl;
	else printf("%d\n",ans);
}