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>
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int maxn=18;
int n;
int a[maxn+5],b[maxn+5];
int dp[maxn+5][(1<<maxn)+5],id[maxn+5];
int p2[maxn+5];
void chm(int &x,int y)
{
	if(x==-1||x>y) x=y;
	return;
}
int count(int num)
{
	int cnt=0;
	for(int i=num;i>0;i-=lowbit(i)) cnt++;
	return cnt;
}
int main()
{ 
	p2[0]=1;
	for(int i=1;i<=maxn;i++) p2[i]=p2[i-1]*2;
	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]);
	int full=(1<<n)-1;
	for(int i=0;i<n;i++) for(int mas=0;mas<=full;mas++) dp[i][mas]=-1;
	for(int i=0;i<n;i++) dp[i][1<<i]=i;
	for(int mas=0;mas<=full;mas++)
	{
		int cnt=count(mas);
		for(int i=0;i<n;i++) 
		{
			if(!(mas&p2[i])) continue;
			if(dp[i][mas]==-1) continue;
			int tot=cnt-1;
			for(int j=0;j<n;j++) if(!(mas&p2[j])) id[j]=++tot;
			int na=(cnt-1)%2==i%2?a[i]:b[i];
			for(int j=0;j<n;j++)
			{
				if(mas&p2[j]) continue;
				int nb=cnt%2==j%2?a[j]:b[j];
				if(na>nb) continue;
				chm(dp[j][mas|p2[j]],dp[i][mas]+id[j]-cnt);
			}
		}
	}
	int ans=-1;
	for(int i=0;i<n;i++) if(dp[i][full]!=-1) chm(ans,dp[i][full]);
	printf("%d\n",ans);
	return 0;
}