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 foi(i,k,n) for(int i = (int)k; i < (int)n; ++i)
using namespace std;
using ll = long long;
using ii = pair<int,int>;
constexpr int MAXN = 18;
constexpr int INF = 1e6;

int memo[1<<MAXN][51];
int a[MAXN], b[MAXN], n;
vector<int> posbits[1<<MAXN];

int dp(int bm, int l) {
  if (bm == 0) return 0;
  int& ans = memo[bm][l];
  if (~ans) return ans;

  ans = INF;
  const vector<int>& v = posbits[bm];
  int pos = n - posbits[bm].size();
  for (int x : v) {
    int izq0 = __builtin_popcount((bm ^ ((1 << n) - 1)) & ~((1 << (1 + x)) - 1));
    int currpos = x + izq0;

    int num = (currpos - pos + izq0) & 1 ? b[x] : a[x];
    if (num >= l) {
      ans = min(ans, currpos - pos + dp(bm ^ (1 << x), num));
    }
  }

  return ans;
}

int main() {
  ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
  cin >> n;
  foi (i,0,n) cin >> a[i];
  foi (i,0,n) cin >> b[i];
  foi (i,0,1<<MAXN) {
    foi (j,0,MAXN) {
      if ((i >> j) & 1) {
        posbits[i].push_back(j);
      }
    }
  }

  memset(memo, -1, sizeof memo);
  cout << (dp((1<<n)-1,0) >= INF ? -1 : dp((1<<n)-1,0)) << '\n';
  return 0;
}