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 <bits/stdc++.h>
#define INF 1000000007
using namespace std;

int n,a[20],b[20],dp[300010][60];

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i=1;i<=n;i++) cin>>b[i];
	for(int i=0;i<=(1<<n)-1;i++) for(int j=1;j<=50;j++) dp[i][j]=INF;
	dp[0][1]=0;
	for(int i=0;i<(1<<n)-1;i++){
		int sum=__builtin_popcount(i);
		for(int j=1;j<=50;j++) if(dp[i][j]!=INF){
			int num=0;
			for(int k=n;k>=1;k--){
				if(i&(1<<(k-1))) num++;
				else{
					int ch=abs(k-(sum+1));
					int cur=0;
					if(ch%2==0) cur=a[k];
					else cur=b[k];
					if(cur>=j){
						dp[i|(1<<(k-1))][cur]=min(dp[i|(1<<(k-1))][cur],dp[i][j]+num);
					}
				}
			}
		}
	}
	int minv=INF;
	for(int i=1;i<=50;i++) minv=min(minv,dp[(1<<n)-1][i]);
	if(minv==INF) minv=-1;
	cout<<minv<<endl;
	return 0;
}