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:
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define mkp make_pair
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
typedef long double ld;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=(1LL<<62);
const int inf=(1<<30);
const int nmax=1e6+50;
const int mod=1e9+7;
using namespace std;
int n,a[20],b[20],f[nmax][55],lm,nr,x,i,j,t,pr,rs=inf;
int main()
{
	//freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n;
	for(i=0;i<n;i++)cin>>a[i];
	for(i=0;i<n;i++)cin>>b[i];
	for(i=0;i<(1<<n);i++)
	{
		nr=__builtin_popcount(i),pr=0;
		if(i)for(t=1;t<=50;t++)f[i][t]=inf;
		for(j=0;j<n;j++)
		{
			if(i&(1<<j))
			{
				pr++;
				x=abs(nr-1-(j+nr-pr));
				if(((nr-1)&1)==(j&1))lm=a[j];
				else lm=b[j];
				for(t=1;t<=lm;t++)if(f[i^(1<<j)][t]!=inf)f[i][lm]=min(f[i][lm],f[i^(1<<j)][t]+x);
				if(i==(1<<n)-1)rs=min(rs,f[i][lm]);
			}
		}
	}
	if(rs==inf)rs=-1;
	cout<<rs<<endl;
	return 0;
}