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 N 200100
using namespace std;
int tmp[18];
bool cmp(int i,int j){
	return tmp[i]<tmp[j];
}
int main(){
	int n,a[2][18],b[18];
	int ans=-1;
	scanf("%d",&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<(1<<n);i++){
		vector<int> s[2];
		for(int j=0;j<n;j++){
			if(i&(1<<j)){
				tmp[j]=a[1][j];
				s[(j&1)^1].push_back(j);
			}
			else{
				tmp[j]=a[0][j];
				s[j&1].push_back(j);
			}
		}
		if(s[1].size()!=n/2) continue;
		for(int j=0;j<2;j++){
			stable_sort(s[j].begin(),s[j].end(),cmp);
		}
		bool flag=false;
		for(int j=1;j<n;j++){
			if(tmp[s[j&1][j>>1]]<tmp[s[(j-1)&1][(j-1)>>1]]) flag=true;
		}
		int tans=0;
		if(!flag){
			for(int j=0;j<n;j++){
				for(int k=j+1;k<n;k++){
					if(s[j&1][j>>1]>s[k&1][k>>1]) tans++;
				}
			}
			if(tans<ans||ans<0) ans=tans;
		}
	}
	printf("%d\n",ans);
	return 0;
}