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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<vector<int>> dp(1 << n, vector<int>(n, INF));
  for (int i = 0; i < n; i++) dp[1 << i][i] = i;
  for (int mask = 0; mask < (1 << n); mask++) for (int last = 0; last < n; last++) {
    if (dp[mask][last] == INF) continue;
    int last_pos = __builtin_popcount(mask) - 1;
    int last_val = ((last_pos & 1) ^ (last & 1) ? b[last] : a[last]);
    int inv = 0;
    for (int add = 0; add < n; add++) {
      if (mask & (1 << add)) continue;
      int add_val = (((last_pos + 1) & 1) ^ (add & 1) ? b[add] : a[add]);
      if (last_val <= add_val) dp[mask | (1 << add)][add] = min(dp[mask | (1 << add)][add], dp[mask][last] + inv);
      inv++;
    }
  }
  int ans = INF;
  for (int i = 0; i < n; i++) ans = min(ans, dp[(1 << n) - 1][i]);
  cout << (ans == INF ? -1 : ans) << '\n';
  return 0;
}