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;

#ifdef iq
  mt19937 rnd(228);
#else
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif

int main() {
#ifdef iq
  freopen("a.in", "r", stdin);
#endif
  ios::sync_with_stdio(0);
  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];
  }
  set <pair <int, int> > q;
  for (int i = 0; i < n; i++) {
    q.insert({b[i], i});
  }
  ll ans = 0;
  while (!q.empty()) {
    int i = q.rbegin()->second;
    q.erase(--q.end());
    if (a[i] == b[i]) {
      continue;
    }
    int delta = b[(i - 1 + n) % n] + b[(i + 1) % n];
    int grab = (b[i] - a[i]) / delta;
    b[i] -= grab * delta;
    ans += grab;
    if (!grab) {
      cout << -1 << '\n';
      return 0;
    }
    q.insert({b[i], i});
  }
  cout << ans << '\n';
}
