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)

/*

  "With age, comes wisdom. With travel, comes understanding.
   Remember that happiness is a way of travel – not a destination"

*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define int            int
#define ld             long double
#define F              first
#define S              second
#define pb             push_back
#define si             set <int>
#define vi             vector <int>
#define pii            pair <int, int>
#define vpi            vector <pii>
#define vpp            vector <pair<int, pii>>
#define mii            map <int, int>
#define mpi            map <pii, int>
#define spi            set <pii>
#define endl           "\n"
#define sz(x)          ((int) x.size())
#define all(p)         p.begin(), p.end()
#define que_max        priority_queue <int>
#define que_min        priority_queue <int, vi, greater<int>>
#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a)       for(auto x : a) cout << x << " "; cout << endl
#define print1(a)      for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y)  for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl

inline int power(int a, int b)
{
     int x = 1;
     while (b)
     {
          if (b & 1) x *= a;
          a *= a;
          b >>= 1;
     }
     return x;
}

typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> Set;

template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
     const char* comma = strchr (names + 1, ',');
     cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}

const int N = (1 << 18) + 5;

int cache[N][52];
pair<int, int> arr[20];
int pos[20];
int n;

int dp(int mask, int val)
{

     if (mask == ((1 << n) - 1))
          return 0;
     int &ans = cache[mask][val];
     if (ans != -1)
          return ans;

     ans = 1e9;
     vector<int> v;

     for (int i = 0; i < n; i++)
     {
          if (mask & (1 << i))
               v.push_back(i + 1);
     }

     int sz = v.size() + 1;

     for (int i = 0; i < n; i++)
     {
          if (mask & (1 << i))
               continue;

          int temp = v.end() - upper_bound(v.begin(), v.end(), i + 1);

          temp += pos[i + 1];
          temp -= sz;

          if (arr[i + 1].first >= val && (abs(sz - pos[i + 1]) % 2) == 0)
               ans = min(ans, temp + dp((mask | (1 << i)), arr[i + 1].first));

          if (arr[i + 1].second >= val && (abs(sz - pos[i + 1]) % 2) != 0)
               ans = min(ans, temp + dp((mask | (1 << i)), arr[i + 1].second));

     }

     return ans;

}

void solve()
{
     memset(cache, -1, sizeof cache);

     cin >> n;
     for (int i = 1; i <= n; i++)
          cin >> arr[i].first;

     for (int i = 1; i <= n; i++)
          cin >> arr[i].second;

     for (int i = 1; i <= n; i++)
          pos[i] = i;

     int ans = dp(0, 0);
     if (ans == 1e9)
          ans = -1;
     cout << ans;
}

int32_t main()
{
     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
     cout << setprecision(9) << fixed;
     clock_t z = clock();
     solve();
     cerr << "Run Time : " << ((double)(clock() - z) / CLOCKS_PER_SEC);

     return 0;
}