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)

#include <bits/stdc++.h>

using namespace std;
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)

typedef long long ll;
typedef pair<ll, ll> P;
const ll INF = 1LL<<60;

ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N;
  cin >> N;
  vector<int> A(N), B(N);
  rep(i, N) cin >> A[i];
  rep(i, N) cin >> B[i];

  int ans = 1 << 30;
  for(int bit = 0; bit < (1 << N); bit++){
    vector<int> S(N);
    vector<int> cnt_flip(N);
    for(int i = 0; i <N; i++){
      if(bit & (1 << i)){
        S[i] = B[i];
      }else{
        S[i] = A[i];
      }
    }
    // S(N)について、転倒数求める
    for(int i = 0; i < N; i++){
      for(int j = 0; j < i; j++){
        if(S[j] > S[i]){
          cnt_flip[i]++;
        }
      }
    }

    bool ok = true;
    for(int i = 0; i < N; i++){
      if(bit & (1 << i)){
        if(cnt_flip[i] % 2 == 0){
          ok = false;
        }
      }else{
        if(cnt_flip[i] % 2 == 1){
          ok = false;
        }
      }
    }
    if(ok){
      int tmp_cnt = 0;
      for(int i = 0; i < N; i++){
          tmp_cnt += cnt_flip[i];
      }
      chmin(ans, tmp_cnt);
    }
  }

  if(ans == 1 << 30){
    cout << -1 << endl;
  }else{
    cout << ans << endl;
  }

  return 0;
}