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;

template<typename T>
void chmin(T &a,T b){
  if(a>b)a=b;
}
struct e{int a,b;};
int ans=1e9,n,a[20],b[20],now,p,q;
e v[20];
stack<int> sta;
#define rep(i,n) for(int i=0;i<(n);i++)

signed main(){
  cin>>n;
  rep(i,n)cin>>a[i];
  rep(i,n)cin>>b[i];
  rep(i,1<<n){
    now=0;
    rep(j,n)v[j]={(i>>j)&1?b[j]:a[j],j};
    rep(j,n){
      int k=j;
      while(k&&v[k].a<v[k-1].a)swap(v[k],v[k-1]),now++,k--;
    }
    rep(j,n)if(((i>>v[j].b)+v[j].b-j)&1)sta.push(v[j].a*1000+j);
    if(sta.size()&1)now=1000,sta.pop();
    while(sta.size()){
      p=sta.top();sta.pop();q=sta.top();sta.pop();
      now+=((p-q)&1?p-q:1000);
    }
    chmin(ans,now);
  }
  cout<<(ans<1000?ans:-1)<<endl;
}