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;

const int N = 200000;

int n;
int a[N + 5], b[N + 5];

#define lst(x) (x == 1 ? n : x - 1)
#define nxt(x) (x == n ? 1 : x + 1)

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);

  long long ans = 0;
  for (bool flag = 1; flag; ) {
    flag = false;
    for (int i = 1; i <= n; ++i) if (b[i] > a[i] && b[i] - a[i] >= b[lst(i)] + b[nxt(i)]) {
      flag = true;
      ans += (b[i] - a[i]) / (b[lst(i)] + b[nxt(i)]);
      b[i] = (b[i] - a[i]) % (b[lst(i)] + b[nxt(i)]) + a[i];
    }
  }

  for (int i = 1; i <= n; ++i) {
    if (a[i] != b[i]) {
      printf("-1\n");
      return 0;
    }
  }
  printf("%lld\n", ans);

  return 0;
}
