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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n + n; i++) {
    cin >> (i < n ? a[i] : b[i - n]);
  }
  priority_queue<pair<int, int>> que;
  for (int i = 0; i < n; i++) {
    que.emplace(b[i], i);
  }
  long long ans = 0;
  while (que.size()) {
    int i = que.top().second;
    que.pop();
    int nbr = b[(i - 1 + n) % n] + b[(i + 1) % n];
    if ((b[i] - a[i]) % nbr == 0) {
      ans += (b[i] - a[i]) / nbr;
      b[i] = a[i];
      continue;
    }
    if (nbr >= b[i]) {
      cout << -1 << endl;
      return 0;
    }
    ans += b[i] / nbr;
    b[i] %= nbr;
    if (b[i] < a[i]) {
      cout << -1 << endl;
      return 0;
    }
    que.emplace(b[i], i);
  }
  cout << ans << endl;
  return 0;
}