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>

#define int long long

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; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  int ans = 0;
  while (true) {
    bool found = false;
    for (int i = 0; i < n; i++) {
      int x = b[(i - 1 + n) % n], &y = b[i], z = b[(i + 1) % n];
      assert(x > 0 && y > 0 && z > 0);
      if (y < (x + z)) {
        continue;
      }
      int l = 0, r = 1e9;
      while (r - l > 1) {
        int mid = (l + r) >> 1;
        if (y - (x + z) * mid >= a[i]) {
          l = mid;
        } else {
          r = mid;
        }
      }
      ans += l;
      y -= l * (x + z);
      found |= (bool)l;
    }
    if (!found) {
      break;
    }
  }
  cout << (a == b ? ans : -1) << '\n';
  return 0;
}

