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;

long long n;
vector<long long> a, b;
priority_queue<pair<int, int>> pq;

long long solve();

int main() {
  cin >> n;
  a.resize(n);
  b.resize(n);
  for(int i = 0; i < n; ++i) cin >> a[i];
  for(int i = 0; i < n; ++i) cin >> b[i];
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long cnt = 0;
  for(int i = 0; i < n; ++i)
    if(b[i] != a[i]) pq.push({b[i] - a[i], i});
  while(!pq.empty()) {
    pair<int, int> now = pq.top();
    pq.pop();
    if(now.first == 0) break;
    long long mnum = b[(now.second + 1) % n] +
                     b[(now.second - 1 + n) % n],
              nowc;
    nowc = (b[now.second] - a[now.second]) / mnum;
    cnt += nowc;
    b[now.second] -= nowc * mnum;
    if(nowc <= 0) return -1;
    if(b[now.second] == a[now.second]) continue;
    pq.push({b[now.second] - a[now.second], now.second});
  }
  return cnt;
}
