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 pb push_back
#define F first
#define S second
#define mkp make_pair
#define pii pair<int,int>
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=2e5+5;
pair<int,int>pre[maxn],after[maxn];
int n;
int vis[maxn],a[maxn],b[maxn];
int solve(int s){
	int cnt=0;
	for(int i=0;i<n;i++)vis[i]=0;
	for(int i=0;i<n;i++){
		if((s)&(1<<i)){
			cnt++;
			pre[i]=mkp(b[i],1);
			after[i]=pre[i];
		}
		else {
			pre[i]=mkp(a[i],0);
			after[i]=pre[i];
		}
	}
	if(cnt&1)return inf;
	cnt=0;
	sort(after,after+n);
	//for(int i=0;i<n;i++)cout<<after[i].first<<" "<<after[i].second<<endl;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(after[j].first==pre[i].first&&(abs(i-j)%2==pre[i].second)&&vis[j]==0)vis[j]=1,cnt++;
		}
	}
//	cout<<cnt<<" "<<n<<endl;
	if(cnt!=n)return inf;
	int ans=0;
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			if(pre[j]<pre[i])ans++;
		}
	}
	return ans;
}
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]);
	int ans=0x3f3f3f3f;
	//solve(6);
	for(int i=0;i<1<<n;i++){
		ans=min(ans,solve(i));
	}
	printf("%d\n",ans==inf?-1:ans);
}