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:
#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> l_l;
typedef pair<int,int> i_i;
const ll mod=1e9+7;
const int inf=1<<30;

int dp[1<<18][51];

int main(){
  int n; cin >> n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  vector<int> b(n);
  rep(i,n) cin >> b[i];
  rep(mask,1<<n)rep(last,51) dp[mask][last]=inf;
  dp[0][0]=0;
  rep(mask,1<<n)rep(last,51)if(dp[mask][last]!=inf){
    vector<i_i> zeros;
    int itr=0;
    rep(i,n){
      if(mask&(1<<i)) continue;
      zeros.emplace_back(i,itr++);
    }
    for(auto v:zeros){
      bool f=abs(__builtin_popcount(mask)-v.first)%2;
      if(f==0&&a[v.first]<last||f==1&&b[v.first]<last) continue;
      chmin(dp[mask|(1<<v.first)][f==0?a[v.first]:b[v.first]],dp[mask][last]+v.second);
    }
  }
  int ans=inf;
  rep(last,51) chmin(ans,dp[(1<<n)-1][last]);
  if (ans==inf) ans=-1;
  cout << ans << endl;
}