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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=20;
const int INF=1061109567;
int n;
int a[N],b[N];
int f[1<<N][N];
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]);
	memset(f,63,sizeof(f));
	f[0][0]=0;
	for(int s=1;s<(1<<n);s++)
	{
		int t=__builtin_popcount(s);
		for(int i=0;i<n;i++)
			if(s&(1<<i))
			{
				int v=(abs(t-(i+1))&1)?b[i]:a[i];
				int num=0;
				for(int j=i+1;j<n;j++)
					if(s&(1<<j)) num++;
				for(int j=0;j<n;j++)
					if((s^(1<<i))==0)
					{
						f[s][i]=min(f[s][i],f[s^(1<<i)][j]+num);
						continue;
					}
					else if((s^(1<<i))&(1<<j))
					{
						int vt=(abs(t-1-(j+1))&1)?b[j]:a[j];
						if(vt<=v) f[s][i]=min(f[s][i],f[s^(1<<i)][j]+num);
					}
			}
	}
	int ans=INF;
	for(int i=0;i<n;i++)
		ans=min(ans,f[(1<<n)-1][i]);
	if(ans>=INF) printf("-1");
	else printf("%d",ans);
	return 0;
}