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 fo(i,a,b) for(int i=(a);i<=(b);++i)
#define rv(i,a,b) for(int i=(a);i>=(b);--i)
#define P pair<int,int>
#define all(x) x.begin(),x.end()
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long LL;
const int M = 1200000;
const int inf = 0x3f3f3f3f;
int n,m,k;
int a[20][2];
int cal(int t[]){
	int ret=0;
	fo(i,0,n-1){
		fo(j,0,i-1){
			if(t[j]>t[i])
				++ret;
		}
	}
	return ret;
}
int main() {
	cin>>n;
	int ans=inf;
	fo(i,0,n-1)
		cin>>a[i][0];
	fo(i,0,n-1)
		cin>>a[i][1];
	int al=(1<<n)-1;
	fo(i,0,al){
		vector<P> v[2];
		fo(j,0,n-1){
			int to=((i>>j)&1);
			v[to^(j&1)].push_back({a[j][to],j});
		}
		if(SZ(v[1])!=n/2 || SZ(v[0])!=(n+1)/2) continue;
		if(!v[0].empty())
			sort(all(v[0]));
		if(!v[1].empty())
			sort(all(v[1]));
		int tar[20];
		int pre=-inf;
		bool bk=0;
		fo(j,0,n-1){
			P cur=v[j&1][j>>1];
			if(cur.first<pre) {
				bk=1;
				break;
			}
			pre=cur.first;
			tar[j]=cur.second;
		}
		if(bk) continue;
		ans=min(ans,cal(tar));
	}
	if(ans==inf) cout<<-1<<endl;
	else cout<<ans<<endl;
	return 0;
}