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>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
int main() {
  int n;
  ll cnt = 0;
  vector<ll> a, b;
  priority_queue<pair<ll, int> > q;
  cin >> n;
  a.resize(n);
  b.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    q.push({b[i], i});
  }
  while (!q.empty()) {
    int x, y;
    x = q.top().first;
    y = q.top().second;
    q.pop();
    if (x != a[y]) {
      if (x - b[(n + y - 1) % n] + b[(n + y + 1) % n] < a[y]) {
        cout << -1 << endl;
        return 0;
      }
      int z = (b[y] - max(a[y], (b[(n + y - 1) % n] + b[(n + y + 1) % n]))) /
              (b[(n + y - 1) % n] + b[(n + y + 1) % n]);
      z = max(z, 1);
      cnt += z;
      b[y] -= z * (b[(n + y - 1) % n] + b[(n + y + 1) % n]);
      q.push({b[y], y});
    }
  }
  cout << cnt << endl;
}