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<long long> A(N), B(N);
  for(auto& a : A) cin >> a;
  for(auto& b : B) cin >> b;
  
  priority_queue<pair<long long, int>> pq;
  for(int i = 0; i < N; i++){
    if(A[i] != B[i]) pq.push(make_pair(B[i], i));
  }
  
  long long turn = 0;
  while(!pq.empty()){
    int i = pq.top().second;
    pq.pop();
    
    long long d = B[(N + i - 1) % N] + B[(i + 1) % N];
    
    if(B[i] - A[i] < d){
      turn = -1;
      break;
    }
    
    turn += (B[i] - A[i]) / d;
    B[i] = A[i] + (B[i] - A[i]) % d;
    if(B[i] > A[i]) pq.push(make_pair(B[i], i));
  }
  
  cout << turn << endl;
  
  return 0;
}