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;

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n, a[200100], b[200100];
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];

  set<pair<int, int>> s;
  for (int i = 0; i < n; i++)
  {
    cin >> b[i];
    s.insert({b[i], i});
  }

  long long ans = 0;
  while (!s.empty())
  {
    int i = s.rbegin()->second;
    s.erase(*s.rbegin());
    if (b[i] <= a[i])
      continue;

    int sum = b[(i + 1) % n] + b[(i + n - 1) % n];
    int op = (b[i] - a[i]) / sum;
    if (op)
    {
      ans += op;
      b[i] -= op * sum;
      s.insert({b[i], i});
    }
    else
    {
      cout << -1 << endl;
      return 0;
    }
  }

  for (int i = 0; i < n; i++)
    if (a[i] != b[i])
    {
      cout << -1 << endl;
      return 0;
    }

  cout << ans << endl;
}
