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>

#define fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;
typedef long long ll;

const int maxn = 55;

int val[20][20];
int N;
int f[1 << 18][maxn];

signed main(void)
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  if (fopen("A.INP", "r")){
    freopen("A.INP", "r", stdin);
    freopen("A.OUT", "w", stdout);
  }
  cin >> N; vector<int> a(N), b(N);
  for (auto & x : a) cin >> x;
  for (auto & x : b) cin >> x;
  for (int i = 0; i < N; ++i){
    for (int j = 0; j < N; ++j){
      val[i][j] = (i % 2 == j % 2) ? a[i] : b[i];
    }
  }
  fill_n(&f[0][0], (1 << 18) * maxn, 1e9 + 5);
  f[0][0] = 0;
  for (int mask = 0; mask < (1 << N); ++mask){
    int i = __builtin_popcount(mask);
    for (int V = 0; V < maxn; ++V){
      int cost = 0;
      for (int j = 0; j < N; ++j){
        if (mask >> j & 1) continue;
        if (val[j][i] >= V){
          f[mask + (1 << j)][val[j][i]] = min(f[mask + (1 << j)][val[j][i]], f[mask][V] + cost);
        }
        ++cost;
      }
    }
  }
  int ret = 1e9;
  for (int i = 0; i < maxn; ++i){
    ret = min(ret, f[(1 << N) - 1][i]);
  }
  if (ret == 1e9) ret = -1;
  cout << ret << '\n';
}