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 <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
int main()
{
  int n;
  cin >> n;
  vector<int> a(n),b(n);
  rep(i,n){
    cin >> a[i];
  }
  rep(i,n){
    cin >> b[i];
  }
  
  const int inf = 1e+9;

  vector<vector<int>> dp((1<<n),vector<int>(51,inf));
  dp[0][1] = 0;
  
  rep(S,(1<<n)){
    int cnt = __builtin_popcount(S);
    rep1(i,50){
      if(dp[S][i]==inf){
	continue;
      }
      int cnt2=0;
      rep(j,n){
	if((S>>j)&1){
	  cnt2++;
	}
	else{
	  if(cnt%2==j%2){
	    if(i<=a[j]){
	      dp[S|(1<<j)][a[j]] = min(dp[S|(1<<j)][a[j]],dp[S][i]+(cnt-cnt2));
	    }
	  }
	  else{
	    if(i<=b[j]){
	      dp[S|(1<<j)][b[j]] = min(dp[S|(1<<j)][b[j]],dp[S][i]+(cnt-cnt2));
	    }
	  }	  
	}
      }
    }
  }

  int res = inf;
  rep1(i,50){
    res = min(res,dp[(1<<n)-1][i]);
  }
  if(res==inf) cout << -1 << "\n";
  else cout << res << "\n";
  
  return 0;
}