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>
using namespace std;

const int N=20;

pair<int,int>a[N];
int b[N],c[N],d[N],vis[N],f[N];

int main() {
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&a[i].first);
	for(int i=0;i<n;i++) scanf("%d",&a[i].second);
	int Ans=1e9;
	for(int i=0;i<(1<<n);i++) {
		for(int j=0;j<n;j++) 
			if(i&(1<<j)) b[j]=a[j].second,f[j]=1;
			else b[j]=a[j].first,f[j]=0;
		for(int j=0;j<n;j++) c[j]=b[j];
		sort(c,c+n);
		memset(vis,0,sizeof vis);
		int flag=0;
		for(int j=0;j<n;j++) {
			int k;
			for(k=0;k<n;k++) 
				if(!vis[k]&&c[k]==b[j]&&((k-j)&1)==f[j]) {
					vis[k]=1;
					d[k]=j;
					break;
				}
			if(k==n) {
				flag=1;
				break;
			}
		}
		if(flag) continue;
		int now=0;
		for(int j=0;j<n;j++)
			for(int k=0;k<j;k++)
				if(d[k]>d[j]) now++;
		Ans=min(Ans,now);
	}
	if(Ans==1e9) puts("-1");
	else cout<<Ans<<endl;
}