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>
#include<cmath>
using namespace std;
int ans=0,lim,cnt=0,dp[55][300000],ct[300005],n,a[2][20],v;
const int inf=1e9;
int main()
{
	scanf("%d",&n);
	lim=(1<<n);
	for(int i=0;i<n;i++)scanf("%d",&a[0][i]);
	for(int i=0;i<n;i++)scanf("%d",&a[1][i]);
	for(int i=0;i<=50;i++)for(int j=0;j<lim;j++)dp[i][j]=inf;
	ct[0]=0;
	for(int i=1;i<lim;i++)
	{
		for(int j=0;j<n;j++)
		{
			if((1<<j)&i)
			{
				ct[i]=ct[i-(1<<j)]+1;
				break;
			}
		}
	}
	dp[0][0]=0;
	for(int i=0;i<lim;i++)
	{
		cnt=0;
		for(int j=n-1;j>=0;j--)
		{
			if(i&(1<<j))
			{
				cnt++;
				continue;
			}
			v=a[abs(j-ct[i])&1][j];
			for(int k=0;k<=v;k++)dp[v][i+(1<<j)]=min(dp[v][i+(1<<j)],dp[k][i]+cnt);
		}
	}
	ans=inf;
	for(int i=0;i<=50;i++)ans=min(ans,dp[i][lim-1]);
	if(ans>=inf)printf("-1\n");
	else printf("%d\n",ans);
	return 0;
}