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=25;
int a[N],b[N],c[N],n;
pair<int,int> d[N];
int cal(int x) {
	int num=0;
	for(int i=0;i<n;i++) {
		if((1<<i)&x) c[i]=b[i],d[i]={b[i],1},num++;
		else c[i]=a[i],d[i]={a[i],0};
	}
	if(num&1) return 1e9;
	sort(c,c+n);
	int ans=0;
	for(int i=n-1;i>=0;i--) {
		bool f=false;
		for(int j=i;j>=0;j--) {
			if(c[i]!=d[j].first||d[j].second!=((i-j)&1)) continue;
			f=true;
			ans+=i-j;
			for(int k=j;k<i;k++) {
				d[k+1].second^=1;
				swap(d[k],d[k+1]);
			}
			break;
		}
		if(!f) return 1e9;
	}
	return ans;
}
int main() {
	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 ans=1e9;
	for(int i=0;i<1<<n;i++) {
		ans=min(ans,cal(i));
	}
	if(ans==1e9) ans=-1;
	printf("%d\n",ans);
	return 0;
}