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 <set>
#include <vector>

int main() {
  int N;
  std::cin >> N;
  std::vector<int> A(N), B(N);
  for (int i = 0; i < N; i++) {
    std::cin >> A.at(i);
  }
  for (int i = 0; i < N; i++) {
    std::cin >> B.at(i);
  }
  std::set<std::pair<int, int>, std::greater<std::pair<int, int>>> s;

  for (int i = 0; i < N; i++) {
    s.emplace(B.at(i), i);
  }

  long long ans = 0;
  while (!s.empty()) {
    int i = s.begin()->second;
    s.erase(s.begin());

    int d = B.at(i) - A.at(i), t = B.at((i + 1) % N) + B.at((N + i - 1) % N);
    if (d != 0 && d < t) {
      ans = -1;
      break;
    }
    ans += d / t;
    B[i] -= d / t * t;

    if (B.at(i) != A.at(i))
      s.emplace(B.at(i), i);
  }

  std::cout << ans;

  return 0;
}