There are N positive integers arranged in a circle.

Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:

* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.



Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.

Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.

Constraints

* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* 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

Print the minimum number of operations required, or `-1` if the objective cannot be achieved.

Examples

Input

3
1 1 1
13 5 7


Output

4


Input

4
1 2 3 4
2 3 4 5


Output

-1


Input

5
5 6 5 2 1
9817 1108 6890 4343 8704


Output

25
#include<iostream>
#include<queue>
using namespace std;
int N_MAX = 200000;
typedef pair<long long,int> P;
int main()
{
  int n,i,j,k;
  long long t,res=0;
  long long a[N_MAX],b[N_MAX];
  P p;
  priority_queue<P> q;
  cin >> n;
  for(i=0;i<n;i++)
    cin >> a[i];
  for(i=0;i<n;i++){
    cin >> b[i];
    p.first=b[i];
    p.second=i;
    q.push(p);
  }
  i=n-1;j=0;k=1;
  while(!q.empty()){
    p=q.top(); q.pop();
    j=p.second;
    i=(j+n-1)%n;
    k=(j+1)%n;
    if(b[j]==a[j])
      continue;
    if(b[j]-b[i]-b[k]<a[j]){
      cout << "-1" << endl;
      return 0;
    }
    t=(b[j]-a[j])/(b[i]+b[k]);
    b[j]-=(b[i]+b[k])*t;
    p.first=b[j];
    q.push(p);
    res+=t;
  }
  for(i=0;i<n;i++)
    if(a[i]!=b[i]){
      cout << "-1" << endl;
      return 0;
    }
  cout << res << endl;
  return 0;
}