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<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long

int main(){
  ll N;
  cin >> N;

  vector<ll> A(N), B(N);
  for(ll i=0; i<N; ++i) cin >> A[i];
  for(ll i=0; i<N; ++i) cin >> B[i];

  ll ans = 1e10;
  ll pos_0 = 0b101010101010101010;

  vector<ll> C(N);
  vector<ll> P(N);

  ll n_operation;
  
  for(ll color=0; !(color>>N); ++color) {
    ll sum_color = 0;
    ll pos_1 = color ^ pos_0;
    for(ll i=0; i<N; ++i) {
      C[i] = ((color>>i)&1 ? B[i] : A[i]);
      P[i] = (pos_1>>i)&1;
      sum_color += ((color>>i)&1);
    }
    if (sum_color&1) continue;

    vector<ll> C_sorted(C);
    sort(C_sorted.begin(), C_sorted.end());
    
    n_operation = 0;
    bool found;
    for(ll i=0; i<N; ++i) {
      found = false;
      ll j=i;
      for(; j<N; ++j) {
        if(C[j]==C_sorted[i] && P[j]==(i&1)) { found = true; break; }
      }

      if (found) {
        n_operation += j-i;
        for(ll k=j; k>i; --k) {
          ll bC = C[k]; C[k]=C[k-1]; C[k-1]=bC;
          ll bP = P[k]; P[k]=P[k-1]; P[k-1]=bP;
        }
        
      } else { break; }

    }

    if(found) ans = min(ans, n_operation);
    
  }

  cout << (ans<1e10 ? ans : -1) << endl;
  return 0;
}