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;
typedef long long ll;

int main() {
  int N;
  cin >> N;
  vector<int> A(N), B(N);
  for (int& i : A) {
    cin >> i;
  }
  for (int& i : B) {
    cin >> i;
  }
  ll res = 0;
  priority_queue<pair<int, int>> Q;
  for (int i = 0; i < N; i++) {
    if (B[i] != A[i]) {
      Q.emplace(B[i], i);
    }
  }
  while (!Q.empty()) {
    int v, p;
    tie(v, p) = Q.top();
    Q.pop();
    int l = B[(p - 1 + N) % N];
    int r = B[(p + 1) % N];
    int w = max(l, r);
    int s = l + r;
    if (A[p] >= w && (B[p] - A[p]) % s == 0) {
      res += (B[p] - A[p]) / s;
      B[p] = A[p];
    } else {
      int x = (v - w) / s + 1;
      B[p] -= x * s;
      res += x;
    }
    if (B[p] < A[p]) {
      return cout << -1 << endl, 0;
    }
    if (B[p] > A[p]) {
      Q.emplace(B[p], p);
    }
  }
  cout << res << endl;

  return 0;
}
