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;

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int&v : a) cin >> v;
  for (int&v : b) cin >> v;
  int ans = 1e9;
  for (int m = 0; m < 1<<n; m++) {
    vector<pair<int,int>> x, y;
    for (int i = 0; i < n; i++) {
      if (m>>i&1) {
	(i%2?x:y).emplace_back(b[i], i);
      } else {
	(i%2?y:x).emplace_back(a[i], i);
      }
      if (x.size() == (n+1)/2) {
	sort(x.begin(), x.end());
	sort(y.begin(), y.end());
	vector<int> v(n), p(n);
	for (int i = 0; i < x.size(); i++)
	  tie(v[i*2], p[i*2]) = x[i];
	for (int i = 0; i < y.size(); i++)
	  tie(v[i*2+1], p[i*2+1]) = y[i];
	int ok = 1;
	for (int i = 1; i < n; i++)
	  if (v[i] < v[i-1]) ok = 0;
	if (!ok) continue;
	int inversions = 0;
	for (int i = 0; i < n; i++)
	  for (int j = 0; j < i; j++)
	    inversions += p[j] > p[i];
	ans = min(ans, inversions);
      }
    }
  }
  if (ans == 1e9) ans = -1;
  cout << ans << endl;
}