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;
  scanf("%d",&N);
  vector<ll> A(N);
  for(int i=0; i<N; i++){
    scanf("%lld",&A[i]);
  }
  priority_queue<pair<int, int>> pq;
  vector<ll> B(N);
  for(int i=0; i<N; i++){
    scanf("%lld",&B[i]);
    pq.push(make_pair(B[i], i));
  }

  ll cnt = 0;
  while (!pq.empty()) {
    pair<int, int> cur = pq.top(); pq.pop();
    int b = cur.second;
    int a = (b-1+N)%N;
    int c = (b+1)%N;
    ll ac = B[a]+B[c];

    if (B[b] == A[b]) continue;
    if (B[b] - ac < A[b]) {
      cout << -1 << endl;
      return 0;
    }

    ll tmp = (B[b]-A[b])/ac;
    B[b] -= ac*tmp;
    cnt += tmp;
    pq.push(make_pair(B[b], b));
  }
  cout << cnt << endl;

  return 0;
}
