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 <stdio.h>
#include <queue>
#include <utility>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, ans = 0, now;
  scanf("%lld", &n);
  ll a[n], b[n];
  for(i = 0; i < n; ++i) scanf("%lld", &a[i]);
  for(i = 0; i < n; ++i) scanf("%lld", &b[i]);
  priority_queue< pair<ll, ll> > q;
  pair<ll, ll> p;
  for(i = 0; i < n; ++i) {
    if(a[i] > b[i]) {
      printf("-1");
      return 0;
    }
    if(a[i] < b[i]) q.push(make_pair(b[i], i));
  }
  while(q.size()) {
    p = q.top(), q.pop();
    now = b[(p.second + 1) % n] + b[(p.second + n - 1) % n];
    if(now > p.first) {
      printf("-1");
      return 0;
    }
    ans += (p.first - a[p.second]) / now;
    b[p.second] = p.first - (p.first - a[p.second]) / now * now;
    if(b[p.second] != a[p.second]) q.push(make_pair(b[p.second], p.second));
  }
  printf("%lld", ans);
  return 0;
}