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;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long
const int INF = 1LL << 60;

signed main(){

  int n; cin >> n;
  vector<int> a(n), b(n);
  for(int i = 0; i < n; i++) cin >> a[i];

  priority_queue<pair<int ,int>> q;
  for(int i = 0; i < n; i++){
    cin >> b[i];
    q.push({b[i], i});
  }

  int ans = 0;
  while(!q.empty()){
    auto p = q.top(); q.pop();
    int idx = p.second;
    int left = (idx - 1 + n) % n, right = (idx + 1) % n;
    int diff = (b[idx] - a[idx]) / (b[left] + b[right]);
    int val = b[idx] - (b[left] + b[right]) * diff;
    if(diff == 0 || val < a[idx]) continue;
    b[idx] = val;
    q.push({val, idx});
    ans += diff;
  }
  if(a != b) ans = -1;
  cout << ans << endl;

  return 0;
}
