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 MX=1e9;
int dp[1000000][52];
int c[1000000][20];
int n;
int a[20];
int b[20];
int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    cin>>n;
    int i,j,k;
    for(i=0 ; i<n ; i++)cin>>a[i];
    for(i=0 ; i<n ; i++)cin>>b[i];
    for(i=0 ; i<=50 ; i++)dp[0][i]=0;
    for(i=1 ; i<(1<<n) ; i++)for(j=0 ; j<=50 ; j++)dp[i][j]=MX;
    for(i=0 ; i<(1<<n) ; i++){
        int k=n;
        for(j=n-1 ; j>=0 ; j--){
            if(((1<<j)&i)==0){
                c[i][j]=k--;
            }
        }
    }
    for(i=1 ; i<(1<<n) ; i++){
        int t=i;
        int s=0;
        while(t){
            s+=t%2;
            t/=2;
        }
        for(k=0 ; k<n ; k++){
            if(((1<<k)&i)==0)continue;
            if(s%2!=k%2){
                dp[i][a[k]]=min(dp[i][a[k]],dp[i^(1<<k)][a[k]]+abs(s-c[i^(1<<k)][k]));
            }
            else{
                dp[i][b[k]]=min(dp[i][b[k]],dp[i^(1<<k)][b[k]]+abs(s-c[i^(1<<k)][k]));
            }
        }
        for(j=1 ; j<=50 ; j++)dp[i][j]=min(dp[i][j],dp[i][j-1]);

    }
    int ans=dp[(1<<n)-1][50];
    if(ans<MX)cout<<ans;
    else cout<<-1;
    return 0;
}