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<bits/stdc++.h>
#include<vector>

using namespace std;

typedef long long ll;

int n;
int a[20],b[20];

struct Node
{
	int face, val;
} c[20];

int d[20];

int solve(int s)
{
	int ret=0;
	for (int i=0;i<n;i++)
	{
		if ((1<<i)&s) // B-side
		{
			c[i+1].face = 1;
			c[i+1].val = b[i+1];
			d[i+1] = b[i+1];
		}
		else // A side
		{
			c[i+1].face = 0;
			c[i+1].val = a[i+1];
			d[i+1] = a[i+1];
		}
	}
	
	sort(d+1,d+n+1);
	
	for (int i=1;i<=n;i++) // 构造答案 
	{
		bool flag = 0;
		for (int j=i;j<=n;j++)
			if ( (c[j].val == d[i]) && (c[j].face == (j-i)%2) )
			{
				flag=1;
				ret+=j-i;
				for (int k=j-1;k>=i;k--)
				{
					swap(c[k],c[k+1]);
					c[k+1].face^=1;
				} 
				break;
			}
		if (!flag) return 2*1e9;
	}
	
	return ret;
}

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	for(int i=1;i<=n;i++)
	{
		cin>>b[i];
	}
	
	int ans=2*1e9;
	for (int i=0;i<(1<<n);i++)
	{
		ans=min(ans,solve(i));
	}
	
	if (ans== 2*1e9) ans=-1;
	printf("%d\n",ans);
}