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;

const int N = 2e5 + 5;

ll a[N], b[N];

int main() {
  int n; scanf("%d", &n);
  priority_queue < pair <ll, ll> > pq;
  for(int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  for(int i = 0; i < n; i++) {
    scanf("%lld", &b[i]);
    if(b[i] > a[i]) pq.push({b[i], i});
  }
  ll moves = 0;
  while(not pq.empty()) {
    auto cur = pq.top(); pq.pop();
    ll have = cur.first;
    int idx = cur.second;
    if(b[idx] != have) continue;
    ll d = b[(idx + n - 1) % n] + b[(idx + 1) % n];
    if(d >= b[idx]) break;
    if((b[idx] - a[idx]) % d == 0) {
      moves += (b[idx] - a[idx]) / d;
      b[idx] = a[idx];
    } else {
      moves += b[idx] / d;
      b[idx] %= d;
    }
    if(b[idx] > a[idx]) pq.push({b[idx], idx});
  }
  for(int i = 0; i < n; i++) {
    if(a[i] != b[i]) moves = -1;
  }
  printf("%lld\n", moves);
}