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;

#define int long long
#define rep(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)x.size())
template<class T>bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,T b){if(a>b){a=b;return 1;}return 0;}

/*
*/

using vi = vector<int>;
using vvi = vector<vi>;
using P = pair<int,int>;

signed main() {
  const int inf = 1LL<<60;
  int n;
  cin >> n;

  vi a(n), b(n);
  for(auto& i : a) cin >> i;
  for(auto& i : b) cin >> i;

  vvi dp(1<<n, vi(55, inf));
  dp[0][0] = 0;
  rep(mask, 0, 1<<n){
    int i = __builtin_popcount(mask);
    vector<P> v;
    int cnt = i;
    rep(j, 0, n) if(!((mask>>j) & 1)) v.emplace_back(j, cnt++);
    rep(m, 0, 55){
      if(dp[mask][m] >= inf) continue;
      for(auto k : v){
        int j = k.first;
        int tmp = j%2 == i%2 ? a[j] : b[j];
        if(tmp < m) continue;
        chmin(dp[mask | (1<<j)][tmp], dp[mask][m]+abs(k.second-i));
      }
    }
  }

  int ans = inf;
  rep(i, 0, 55){
    chmin(ans, dp[(1<<n)-1][i]);
  }
  if(ans < inf) cout << ans << endl;
  else cout << -1 << endl;

  return 0;
}