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;
using ll=long long;
const int MOD=(int)1e9+7;
const int INF=(int)1e9;
const ll LINF=(ll)1e18;

int main(){
    int n;
    cin>>n;
    int a[n],b[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<n;i++){
        cin>>b[i];
    }
    int dp[1<<n][55];
    fill(dp[0],dp[1<<n],INF);
    dp[0][0]=0;
    for(int s=0;s<(1<<n)-1;s++){
        int cnt=0;
        for(int i=0;i<n;i++){
            if((s>>i)&1)cnt++;
        }
        for(int mx=0;mx<=50;mx++){
            if(dp[s][mx]==INF)continue;
            int pos=cnt;
            for(int i=0;i<n;i++){
                if((s>>i)&1)continue;
                int num=((cnt-i)&1)?b[i]:a[i];
                if(mx<=num){
                    dp[s+(1<<i)][num]=min(dp[s+(1<<i)][num],dp[s][mx]+abs(pos-cnt));
                }
                pos++;
            }
        }
    }
    int ans=INF;
    for(int i=1;i<=50;i++){
        ans=min(dp[(1<<n)-1][i],ans);
    }
    cout<<(ans==INF?-1:ans)<<endl;
}