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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n; cin >> n;
  V<> a(n); for (auto&& e : a) cin >> e;
  V<> b(n); for (auto&& e : b) cin >> e;
  priority_queue< pair<int, int> > pque;
  for (int i = 0; i < n; ++i) {
    if (a[i] > b[i]) return cout << -1 << '\n', 0;
    if (b[i] > a[i]) {
      pque.emplace(b[i], i);
    }
  }
  auto _ = [&](int i) -> int { return (n + i) % n; };

  lint res = 0;
  while (!pque.empty()) {
    int x, i;
    tie(x, i) = pque.top(); pque.pop();
    int s = b[_(i - 1)] + b[_(i + 1)];
    int q = (b[i] - a[i]) / s;
    if (!q) return cout << -1 << '\n', 0;
    b[i] -= q * s;
    res += q;
    if (b[i] > a[i]) {
      pque.emplace(b[i], i);
    }
  }
  cout << res << '\n';
}