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 inf 0x3f3f3f3f
using namespace std;
const int N=20;
int n,a[N],b[N];
struct node
{
    int x,p;
    node(int x=0,int p=0):x(x),p(p){}
}c[N];
int d[N];
int solve(int x)
{
    for(int i=0;i<n;i++)
        if(x>>i&1)
            c[i+1]=node(b[i+1],1),d[i+1]=b[i+1];
        else
            c[i+1]=node(a[i+1],0),d[i+1]=a[i+1];
    sort(d+1,d+1+n);
    int ans=0;
    for(int i=n;i>=1;i--)
    {
        bool flag=false;
        for(int j=i;j>=1;j--)
            if(c[j].x==d[i]&&c[j].p==(i-j&1))
        {
            flag=true;
            ans+=i-j;
            for(int k=j;k<i;k++)
            {
                c[k+1].p^=1;
                swap(c[k],c[k+1]);
            }
            break;
        }
        if(!flag) return inf;
    }
    return ans;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=n;i++) scanf("%d",&b[i]);
    int ans=inf;
    for(int i=0;i<1<<n;i++)
        ans=min(ans,solve(i));
    if(ans==inf) ans=-1;
    printf("%d\n",ans);
}