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;

#define rep(i, a, b) for (int i = (a); i <= (b); i++)
typedef long long ll;
const int N = 2e5 + 100;

int n, a[N], b[N]; ll ans;
struct nod {
  int d, id;
  inline bool operator < (const nod &rhs) const {
    return d < rhs.d;
  }
} ;
priority_queue <nod> Q;

inline int ID(int x) {
  if (x > n) return x - n;
  if (x < 1) return x + n;
  return x;
}

int main() {
  scanf("%d", &n); rep(i, 1, n) scanf("%d", &a[i]);
  rep(i, 1, n) { scanf("%d", &b[i]); Q.push((nod) {b[i], i}); }
  while (!Q.empty()) {
    int p = Q.top().id, lr = b[ID(p + 1)] + b[ID(p - 1)];
    int c = (b[p] - a[p]) / lr; Q.pop();
    if (a[p] == b[p]) continue ; if (c <= 0) { puts("-1"); return 0; }
    b[p] -= lr * c; Q.push((nod) {b[p], p}); ans += c;
  }
  printf("%lld\n", ans);
  return 0;
}