Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize("O3")
#pragma GCC target("sse4")

#include <bits/stdc++.h>
#define gc getchar_unlocked()
#define pc(x) putchar_unlocked(x)
template<typename T> void scan(T &x){x = 0;bool _=0;T c=gc;_=c==45;c=_?gc:c;while(c<48||c>57)c=gc;for(;c<48||c>57;c=gc);for(;c>47&&c<58;c=gc)x=(x<<3)+(x<<1)+(c&15);x=_?-x:x;}
template<typename T> void printn(T n){bool _=0;_=n<0;n=_?-n:n;char snum[65];int i=0;do{snum[i++]=n%10+48;n/= 10;}while(n);--i;if (_)pc(45);while(i>=0)pc(snum[i--]);}
template<typename First, typename ... Ints> void scan(First &arg, Ints&... rest){scan(arg);scan(rest...);}
template<typename T> void print(T n){printn(n);pc(10);}
template<typename First, typename ... Ints> void print(First arg, Ints... rest){printn(arg);pc(32);print(rest...);}

using namespace std;
int n;
map<vector<int>, int> dp;

bool check(vector<int> &v){
    int l = 0;
    for(int i = 0; i < n; i++){
        if(v[i] < l)
            return 0;
        l = v[i];
    }
    return 1;
}

int go(vector<int> v){
    /*
    cout << "go ";
    for(int i: v)
        cout << i << ' ' ;
    cout << endl;
     */
    if(dp[v])
        return dp[v];
    if(check(v))
        return dp[v] = 0;
    int ans = 1e9;
    dp[v] = ans;
    for(int i = 0; i < n-1; i++){
        vector<int> nx = v;
        swap(nx[i], nx[i+n+1]);
        swap(nx[i+1], nx[i+n]);
        ans = min(ans, go(nx)+1);
    }
    return dp[v] = ans;
}

int main(){
    vector<int> pos;
    scan(n);
    for(int i = 0,a; i < n*2; i++){
        scan(a);
        pos.emplace_back(a);
    }
    int ans = go(pos);
    if(ans == 1e9)
        puts("-1");
    else
        print(ans);
    
    return 0;
}