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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
int a[20][2],f[300010][20];
int main()
{
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	int n=read();
	for(int i=0;i<n;i++)a[i][0]=read();
	for(int i=0;i<n;i++)a[i][1]=read();
	memset(f,-1,sizeof(f));
	for(int i=0;i<n;i++)f[1<<i][i]=i;
	for(int i=1;i<(1<<n);i++)
	{
		int tot=0,sum=0;
		for(int j=0;j<n;j++)if(i>>j&1)tot++;
		for(int j=0;j<n;j++)
		{
			if(i>>j&1)
			{
				if(f[i][j]==-1)continue;
				int wy=(tot+2*sum-j-1)&1,cnt=0;
				for(int k=0;k<n;k++)
				{
					if(i>>k&1)continue;
					cnt++;int ul=(tot+2*cnt-k-2)&1;
					if(a[j][wy]>a[k][ul])continue;
					int uf=f[i][j]+cnt-1;
					if(f[i+(1<<k)][k]==-1)f[i+(1<<k)][k]=uf;
					else f[i+(1<<k)][k]=min(f[i+(1<<k)][k],uf);
				}
			}
			else sum++;
		}
	}
	int ans=-1;
	for(int i=0;i<n;i++)
	{
		if(f[(1<<n)-1][i]!=-1)
		{
			if(ans==-1)ans=f[(1<<n)-1][i];
			else ans=min(ans,f[(1<<n)-1][i]);
		}
	}pr2(ans);
	return 0;
}