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>
using namespace std;
int main(){
int N;cin>>N;
  vector<int64_t>A(N);
  vector<int64_t>B(N);
  for(int i=0;i<N;i++)
    cin>>A.at(i);
  for(int i=0;i<N;i++)
    cin>>B.at(i);
  bool update=true;
  int64_t ans=0;
  while(update){
  update=false;
    for(int i=0;i<N;i++){
    int64_t a=B[(N+i-1)%N];
      int64_t b=B[i];
      int64_t c=B[(N+i+1)%N];
      if(a+c<b && A[i]<=b-a-c){
      int64_t d=(b-A.at(i))/(a+c);
        ans+=d;update=true;
        B.at(i)-=d*(a+c);
      }
    }
  
  }bool x=true;
  for(int i=0;i<N;i++)
    if(A[i]!=B[i]){x=false;break;}
  if(!x)
    cout<<-1<<endl;
  else
    cout<<ans<<endl;
return 0;
}