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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define PI 3.141592653589793
#define rep(i, n) for (int i = 0; i < (n); i++)
#define REP(i, a, n) for (int i = a; i < (n); i++)
#define rrep(i, n, k) for (int i = (n); i >= (k); i--);
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
template<class T> istream &operator>>(istream&is,vector<T>&v){for(auto &elemnt:v)is>>elemnt;return is;}
template<class T,class U> istream &operator>>(istream&is,pair<T,U>&p){is>>p.first>>p.second;return is;}
template<class T>vector<T> make_vector(size_t a){return vector<T>(a);}
template<class T, class... Ts>auto make_vector(size_t a, Ts... ts){return vector<decltype(make_vector<T>(ts...))>(a, make_vector<T>(ts...));}
const int inf = 6e18;
const int mod = 1e9 + 7;

signed main() {
    int N;
    cin >> N;
    vector<int> A(N), B(N);
    cin >> A >> B;
    auto dp = vector<vector<int>>(1 << N, vector<int>(N, inf));
    for (int i = 0; i < N; i++) {
        dp[1 << i][i] = i;
    }
    for (int i = 1; i < (1 << N); i++) {
        int now = __builtin_popcount(i);
        for (int j = 0; j < N; j++) {
            if (~i & (1 << j)) continue; // 0 continue
            int add = 0;
            for (int k = 0; k < N; k++) {
                if (i & (1 << k)) continue; // 1 continue
                int before = abs(j - (now - 1)) % 2 == 0 ? A[j] : B[j];
                int select = abs(k - now) % 2 == 0 ? A[k] : B[k];
                if (before <= select) dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + add);
                ++add;
            }
        }
    }
    int ans = inf;
    for (int i = 0; i < N; i++) ans = min(ans, dp[(1 << N) - 1][i]);
    cout << (ans == inf ? -1 : ans) << endl;
}