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;

int a[20][2],n,f[262144][19][2];

int F(int i,int j,int ok)
{
    if(f[i][j][ok]!=-1) return f[i][j][ok];
    else if(i==(1<<n)-1) return 0;
    else
    {
        int res=1e9,sl=0,t=0;
        for(int id=1;id<=n;id++)
        {
            if(((i>>(id-1))&1)==1) sl++;
        }
        //cout<<i<<" "<<j<<" "<<ok<<" "<<sl<<'\n';
        for(int id=1;id<=n;id++)
        {
            if(((i>>(id-1))&1)==0)
            {
                //cout<<id<<'\n';
                int okms=abs(sl-id+1)%2;
                //cout<<id<<" "<<okms<<" "<<abs(id-sl)<<'\n';
                if(a[id][okms]>=a[j][ok]) res=min(res,F(i+(1<<(id-1)),id,okms)+abs(id-t-1));
            }
            else t++;
        }
        f[i][j][ok]=res;
        return res;
    }
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i][0];
    }
    for(int i=1;i<=n;i++)
    {
        cin>>a[i][1];
    }
    memset(f,-1,sizeof(f));
    int ans=F(0,0,0);
    if(ans==1e9) ans=-1;
    cout<<ans;
}