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;

const int MAXN = 18;

int n;
int a[MAXN][2], b[MAXN];
int f[1 << MAXN][55];

int main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL);

  cin >> n;
  for(int j = 0; j < 2; ++j) {
    for(int i = 0; i < n; ++i) cin >> a[i][j];
  }

  memset(f, 60, sizeof f);

  f[0][0] = 0;
  for(int mask = 0; mask + 1 < (1 << MAXN); ++mask) {
    int cur = __builtin_popcount(mask);
    int current_cost = 0;
    int cst[MAXN];
    for(int i = 0; i < n; ++i) {
      cst[i] = current_cost;
      if(!(mask & (1 << i))) {
        ++current_cost;
      }
    }
    for(int j = 0; j <= 50; ++j) {
      for(int i = 0; i < n; ++i) {
        if(!(mask & (1 << i))) {
          int dst = abs(i - cur) % 2;
          if(a[i][dst] >= j) {
            f[mask ^ (1 << i)][a[i][dst]] = min(f[mask ^ (1 << i)][a[i][dst]], f[mask][j] + cst[i]);
          }
        }
      }
    }
  }

  int ans = (int) 1e9;
  for(int i = 0; i <= 50; ++i) {
    ans = min(ans, f[(1 << n) - 1][i]);
  }

  if(ans > 2000000) {
    cout << -1;
  }
  else cout << ans;

  return 0;
}