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 long long ll;
typedef pair<int, int> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int N = 18;
int a[2][N];
int dp[1<<N][N];

int main(){
  fastIO;
  int n;
  cin >> n;
  for(int i = 0 ;i < n; i ++ ){
    cin >> a[i%2][i];
  }
  for(int i = 0 ; i < n; i ++ ){
    cin >> a[1-i%2][i];
  }
  for(int i = 0 ; i < (1 << n); i ++ ){
    for(int j = 0 ; j < n ; j ++ )
      dp[i][j] = (int)1e9;
  }
  for(int i = 0 ; i < n ; i ++ ){
    dp[(1<<i)][i] = i;
  }
  int p;
  int d, las;
  int ct;
  for(int i = 1; i < (1 << n) ; i ++ ){
    p = __builtin_popcount(i);
    for(int l = 0 ; l < n; l ++ ){
      if(dp[i][l] == (int)1e9) continue;
      ct = 0;
      for(int j = 0 ; j < n; j ++ ){
        if(!(i & (1 << j))){
          if(a[(p-1)%2][l] <= a[p%2][j]){
            dp[(i|(1<<j))][j] = min(dp[i|(1<<j)][j],dp[i][l]+ct);
          }
          ct++;
        }
      }
    }
  }
  int res = (int)1e9;
  for(int i = 0; i < n; i ++ )
    res = min(res, dp[(1<<n)-1][i]);
  if(res != (int)1e9)cout << res << "\n";
  else cout << -1 << "\n";
  return 0;
}