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 <bits/stdc++.h>
#define ll long long
using namespace std;

int main(){
  int n;
  cin>>n;
  ll ans=0;
  ll vec[n],vecb[n];
  for(int i=0;i<n;i++){
    cin>>vec[i];
  }
  for(int i=0;i<n;i++){
    cin>>vecb[i];
    if(vecb[i]<vec[i]){
      cout<<-1<<endl;
      return 0;
    }
  }
  for(int q=0;q<60;q++){
  for(int i=0;i<n;i++){
    if(vecb[i]>vecb[(i+1)%n]&&vecb[i]>vecb[(i+n-1)%n]&&vecb[i]!=vec[i]){
      if(vecb[i]<vec[i]+vecb[(i+1)%n]+vecb[(i+n-1)%n]){
        cout<<-1<<endl;
        return 0;
      }
      ll x=(vecb[i]-max(vecb[(i+1)%n],vecb[(i+n-1)%n])+vecb[(i+1)%n]+vecb[(i+n-1)%n]-1)/(vecb[(i+1)%n]+vecb[(i+n-1)%n]);
      x=min(x,(vecb[i]-vec[i]+vecb[(i+1)%n]+vecb[(i+n-1)%n]-1)/(vecb[(i+1)%n]+vecb[(i+n-1)%n]));
      vecb[i]-=(vecb[(i+1)%n]+vecb[(i+n-1)%n])*x;
      ans+=x;
    }
  }
  }
  
  for(int i=0;i<n;i++){
    if(vecb[i]!=vec[i]){
      cout<<-1<<endl;
      return 0;
    }
  }
  cout<<ans<<endl;
}