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;
using ll = long long;

#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >= 0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)

#define INF 1e9

int N;
vector<int> A, B;
int ans = INF;

int main(){
  cin >> N;
  A.resize(N), B.resize(N);
  REP(i, N){
    cin >> A[i];
  }
  REP(i, N){
    cin >> B[i];
  }
  REP(b, 1<<N){
    vector<int> retu;
    int cntB = 0;
    REP(i, N){
      if(b >> i & 1){
        retu.push_back(A[i]);
      }
      else{
        cntB++;
        retu.push_back(B[i]);
      }
    }
    if(cntB %2) continue;
    sort(retu.begin(), retu.end());
    vector<int> homon(N, false), hcnt(N, 0);
    int cost = 0;
    bool flag = true;
    REP(i, N){
      int minc = INF, idx = -1;
      REP(j, N){
        if(homon[j] == true) continue;
        if(b >> i & 1){
          if(abs(i-j) % 2 == 0 && A[j] == retu[i]){
            idx = j;
            break;
          }
        }
        else{
          if(abs(i-j) % 2 == 1 && B[j] == retu[i]){
            idx =j;
            break;
          }
        }
      }
      if(idx == -1){
        flag = false;
        break;
      }
      else{
        cost +=hcnt[idx] + abs(i-(idx+hcnt[idx]));
        homon[idx] = true;
        REP(k, idx){
          hcnt[k]++;
        }
      }
    }
    if(flag){
      ans = min(cost, ans);
    }
  }
  if(ans == INF){
    cout << -1 << endl;
  }
  else{
    cout << ans/2 << endl;
  }
  return 0;
}