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: Wrong Answer)

/*
    author:  Maksim1744
    created: 18.01.2020 15:26:08
*/

#include <bits/stdc++.h>

using namespace std;

#define ll   long long
#define ld   long double

#define mp   make_pair
#define pb   push_back
#define eb   emplace_back

#define sum(a)     ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)    (*min_element((a).begin(), (a).end()))
#define maxe(a)    (*max_element((a).begin(), (a).end()))
#define mini(a)    ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)    ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())

template<typename T>             vector<T>& operator--            (vector<T>& v){for (auto& i : v) --i;            return  v;}
template<typename T>             vector<T>& operator++            (vector<T>& v){for (auto& i : v) ++i;            return  v;}
template<typename T>             istream& operator>>(istream& is,  vector<T>& v){for (auto& i : v) is >> i;        return is;}
template<typename T>             ostream& operator<<(ostream& os,  vector<T>& v){for (auto& i : v) os << i << ' '; return os;}
template<typename T, typename U> istream& operator>>(istream& is, pair<T, U>& p){is >> p.first >> p.second;        return is;}
template<typename T, typename U> ostream& operator<<(ostream& os, pair<T, U>& p){os << p.first << ' ' << p.second; return os;}
template<typename T, typename U> pair<T,U> operator-(pair<T,U> a, pair<T,U> b){return mp(a.first-b.first, a.second-b.second);}
template<typename T, typename U> pair<T,U> operator+(pair<T,U> a, pair<T,U> b){return mp(a.first+b.first, a.second+b.second);}
template<typename T, typename U> void umin(T& a, U b){if (a > b) a = b;}
template<typename T, typename U> void umax(T& a, U b){if (a < b) a = b;}

#ifdef HOME
#define TAG_LENGTH       25
#define LR_LEFT          left
#define LR_RIGHT         right
#define LR_VALUE         value
#define LR_SECOND_VALUE  mn
#include "C:/C++ libs/print.cpp"
#else
#define showl            42;
#define shows            42;
#define show(...)        42;
#define showm(...)       42;
#define showt(...)       42;
#define printTree(...)   42;
#define printGraph(...)  42;
#define printLRTree(...) 42;
#define printMatrix(...) 42;
#define printWGraph(...) 42;
#define debug     if (false)
#endif

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    cin >> a >> b;
    int ans = 1e9;
    for (int i = 0; i < (1 << n); ++i) {
        vector<int> v(n, false);
        for (int j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                v[j] = true;
            }
        }
        vector<int> res;
        for (int j = 0; j < n; ++j) {
            if (v[j]) res.pb(b[j]);
            else res.pb(a[j]);
        }
        sort(res.begin(), res.end());
        vector<int> has;
        for (int j = 0; j < n; ++j) {
            int k = a[j];
            if (v[j]) {
                k = b[j];
            }
            has.pb(k);
        }
        int sm = 0;
        bool ok = true;
        for (int i = 0; i < n; ++i) {
            ok = false;
            for (int j = 0; j < n; ++j) {
                if (has[i] == res[j] && ((i + v[i]) % 2 == j % 2)) {
                    res[j] = -1;
                    sm += abs(i - j);
                    ok = true;
                    break;
                }
            }
            if (!ok) break;
        }
        if (!ok) continue;
        if (sm % 2 == 1) continue;
        ans = min(ans, sm / 2);
    }
    if (ans == 1e9) ans = -1;
    cout << ans << endl;

    return 0;
}