Problem: 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.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  int N; cin >> N;
  vector<int> A(N), B(N);
  rep(i, N) cin >> A[i];
  rep(i, N) cin >> B[i];
  const int INF = 1e9 + 10;
  int ans = INF;

  rep(k, 1<<N) {
    bitset<18> bit(k);
    if (bit.count() % 2 == 1) continue;
    vector<pair<int, int>> C;
    vector<int> S(N);
    int kans = 0;
    rep(i, N) {
      if (bit[i] == 0) C.emplace_back(A[i],0);
      else C.emplace_back(B[i], 1);
      S[i] = C[i].first;
    }
    sort(S.begin(), S.end());

    for (int i = N - 1; i >= 0; i--) {
      bool check = false;
      for (int j = i; j >= 0; j--) {
        if (S[i] != C[j].first) continue;
        if ((i - j + C[j].second) % 2 == 1) continue;
        kans += i - j;
        C.erase(C.begin() + j);
        for (int x = j; x < C.size(); x++) {
          C[x].second ^= 1;
        }
        check = true;
        break;
      }
      if (!check) {
        kans = INF;
        break;
      }
    }
    ans = min(ans, kans);
  }

  if(ans == INF) ans = -1;
  cout << ans << endl;
}