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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n;
ll a[300000], b[300000];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 0; i < n; i++)
    cin >> b[i];
  priority_queue<pair<ll, int>> pq;
  for (int i = 0; i < n; i++)
    if (b[i] > a[i])
      pq.push({b[i], i});
  ll o = 0;
  while (!pq.empty()) {
    auto p = pq.top();
    pq.pop();
    int i = p.second;
    ll c = (b[i] - a[i]) / (b[(i + 1) % n] + b[(i - 1 + n) % n]);
    o += c;
    b[i] -= c * (b[(i + 1) % n] + b[(i - 1 + n) % n]);
    if (c == 0 || b[i] < a[i]) {
      cout << -1 << "\n";
      return 0;
    }
    if (b[i] > a[i])
      pq.push({b[i], i});
  }
  cout << o << "\n";
}
