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;
int dp[1<<18][50];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n][2];
  for (int i = 0; i < n; ++i) cin >> a[i][0], a[i][0]--;
  for (int i = 0; i < n; ++i) cin >> a[i][1], a[i][1]--;
  for (int p = 1; p < (1<<n); ++p) {
    for (int x = 0; x < 50; ++x) dp[p][x] = 2e9;
    int pos = __builtin_popcount(p)-1;
    int xc = pos+1;
    for (int i = 0; i < n; ++i) {
      if ((1<<i) & p) {
        xc--;
        int ov = a[i][(pos^i)&1];
        for (int x = 0; x <= ov; ++x) {
          dp[p][ov] = min(dp[p][ov], dp[p^(1<<i)][x]+xc);
        }
      }
    }
  }
  int res = 2e9;
  for (int x = 0; x < 50; ++x) {
    res = min(res, dp[(1<<n)-1][x]);
  }
  if (res == 2e9) res = -1;
  cout << res << endl;
}