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 vector<int> vint;
#define FOR(i,x,y) for(int i=(int)x; i<(int)y; ++i)
#define REP(i,y) FOR(i, 0, y)
constexpr int INF = 1<<30;
int N;
vector<vint> X(18, vint(2));
vint V_(18), V(18), Vb(18);
inline int search(int F){
  if(__builtin_popcount(F)&1) return INF;
  REP(i,N){
    V[i] = X[i][F%2];
    V_[i] = V[i];
    Vb[i] = F%2;
    F>>=1;
  }
  sort(V.begin(), V.begin()+N);
  int ans = 0;
  REP(i,N){
    FOR(j,i,N){
      if(!(V[i]==V_[j] && (j-i+Vb[j])%2==0)) continue;
      ans += j-i;
      for(int k=j-1; k>=i; --k){
        V_[k+1] = V_[k];
        Vb[k+1] = Vb[k]+1;
      }
      goto END;
    }
    return INF;
END:
    continue;
  }
  return ans;
}
int main(){
  cin >> N;
  REP(i,N) cin >> X[i][0];
  REP(i,N) cin >> X[i][1];
  int ans=INF;
  REP(i,1<<N) ans = min(ans, search(i));
  cout << (ans==INF?-1:ans) << endl;
  return 0;
}