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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; ++i)
#define all(a) (a).begin(), (a).end()

struct Card {
  int i, A, B;
  int val(int j) { return (i - j) % 2 == 0 ? A : B; }
};

int main() {
  int N;
  cin >> N;

  vector<Card> C(N);
  rep(i, N) C[i].i = i;
  rep(i, N) cin >> C[i].A;
  rep(i, N) cin >> C[i].B;

  const int INF = 1e9;
  vector<vector<int>> dp((1 << N), vector<int>(N, INF));

  rep(i, N) dp[0][i] = 0;

  rep(s, (1 << N)) {
    int c = __builtin_popcount(s);
    rep(i, N) {
      if (s && !((s >> i) & 1)) continue;
      if (dp[s][i] == INF) continue;

      int inv = -1;
      rep(j, N) {
        if ((s >> j) & 1) continue;
        ++inv;
        if (c && C[i].val(c - 1) > C[j].val(c)) continue;

        dp[s | (1 << j)][j] = min(dp[s | (1 << j)][j], dp[s][i] + inv);
      }
    }
  }

  int ans = *min_element(all(dp[(1 << N) - 1]));
  if (ans == INF) ans = -1;

  cout << ans << endl;

  return 0;
}