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 <utility>
#include <cstdio>
#define llint long long
#define mod 1000000007
#define inf 1e18

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[2][20];
llint dp[1<<18][55];
llint pop[1<<18];

int main(void)
{
  ios::sync_with_stdio(0);
  cin.tie(0);

  cin >> n;
  for(int i = 0; i < n; i++) cin >> a[0][i];
  for(int i = 0; i < n; i++) cin >> a[1][i];
  for(int i = 1; i < (1<<18); i++) pop[i] = pop[i&(i-1)] + 1;

  llint N = 1<<n;
  for(int i = 0; i < N; i++){
    for(int j = 0; j <= 50; j++){
      dp[i][j] = inf;
    }
  }
  dp[0][0] = 0;

  for(int i = 0; i < N; i++){
    for(int j = 0; j <= 50; j++){
      llint cnt = 0;
      for(int k = 0; k < n; k++){
        if(i & (1<<k)) continue;
        cnt++;
        llint d = abs(pop[i]-k), v = a[d%2][k];
        if(v < j) continue;
        dp[i|(1<<k)][v] = min(dp[i|(1<<k)][v], dp[i][j] + cnt-1);
      }
    }
  }

  /*for(int i = 0; i < N; i++){
    for(int j = 0; j < 5; j++){
      cout << dp[i][j] << " ";
    }
    cout << endl;
  }*/

  llint ans = inf;
  for(int j = 0; j <= 50; j++) ans = min(ans, dp[N-1][j]);
  if(ans > inf/2) ans = -1;
  cout << ans << endl;

  return 0;
}