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 <cstdio>
#include <algorithm>
using namespace std;

int x,y,pi[23][2],pt[2][23],pl[23],sum=2000000000;

bool bj1(int a,int b)
{
	return pi[a][1]<pi[b][1]||(pi[a][1]==pi[b][1]&&a<b);
}

bool bj2(int a,int b)
{
	return pi[a][0]<pi[b][0]||(pi[a][0]==pi[b][0]&&a<b);
}

int suan(int a,int b)
{
	int pp[2][23];
	for(int i=1; i<=b; i++) pp[0][i]=pt[0][i];
	for(int i=1; i<=a; i++) pp[1][i]=pt[1][i];
	sort(pp[1]+1,pp[1]+1+a,bj1);
	sort(pp[0]+1,pp[0]+1+b,bj2);
	for(int i=1; i<=a+b; i++) pl[i]=pp[i%2][(i-1)/2+1];
	for(int i=1; i<a+b; i++) if(pi[pl[i]][i%2]>pi[pl[i+1]][(i+1)%2]) return sum;
	int p=0;
	for(int i=1; i<=a+b; i++)
	{
		p+=pl[i]-i;
		for(int j=1; j<i; j++) if(pl[j]>pl[i]) p++;
	}
	return p;
}

void sou(int k,int a,int b)
{
	if(a==0&&b==0)
	{
		sum=min(sum,suan(x,y));
		return;
	}
	if(a!=0)
	{
		pt[1][a]=k;
		sou(k+1,a-1,b);
	}
	if(b!=0)
	{
		pt[0][b]=k;
		sou(k+1,a,b-1);
	}
	return;
}

int main()
{
	int n; scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%d",&pi[i][i%2]);
	for(int i=1; i<=n; i++) scanf("%d",&pi[i][1-i%2]);
	y=n/2; if(n%2==1) x=y+1; else x=y;
	sou(1,x,y);
	if(sum==2000000000) printf("-1");
	else printf("%d",sum);
	return 0;
}