Problem: 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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
const int N=20,INF=1e9;
int a1[N],a2[N],n,c[N],d[N];
bool b[N];
int slo() {
	for(int i=1;i<=n;i++)
		if(b[i]==1)
			c[i]=a2[i];
		else
			c[i]=a1[i];
	
	for(int i=1;i<=n;i++)
		d[i]=c[i];

	sort(c+1,c+1+n);
	
	int ans=0;
	for(int i=1;i<=n;i++) {
		bool flag=1;
		for(int j=i;j<=n;j++) {
			if(d[j]==c[i] && (j-i+b[j])%2==0) {
				flag=0;
				ans+=j-i;
				for(int k=j;k>i;k--) {
					b[k]=!b[k-1];
					d[k]=d[k-1];
				}
				break;
			}
		}
		if(flag)return -1;
	}
	return ans;
}
int main(){
	
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a1[i];
	for(int i=1;i<=n;i++)
		cin>>a2[i];
	int s=(1<<n)-1;
	int ans=INF;
	for(int i=0;i<=s;i++) {	//1代表翻面 
//		i=6;
		int fl=0;
		for(int j=0;j<n;j++)
			if(i&(1<<j))
				b[j+1]=1,fl++;
			else 
				b[j+1]=0;
		if(fl&1)continue;
		int al=slo();
		if(al!=-1)
			ans=min(ans,al);
	}
	if(ans!=INF)
		printf("%d\n",ans);
	else
		printf("%d\n",-1);
	return 0;
}