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;
using ll = long long;
int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin >> n;
  vector<ll> a(n),b(n);
  priority_queue<pair<ll,ll>> q;
  for(ll i = 0; i < n; ++i) cin >> a[i];
  for(ll i = 0; i < n; ++i){
    cin >> b[i];
    q.emplace(b[i],i);
  }
  ll ans = 0,val,idx;
  while(!q.empty()){
    auto p = q.top();
    q.pop();
    tie(val,idx) = p;
    if(b[idx] < a[idx]){
      cout << -1;
      return 0;
    }
    if(b[idx] != val or b[idx] == a[idx]) continue;
    ll val2 = b[(idx+1)%n]+b[(idx+n-1)%n];
    if(val < val2+a[idx]){
      cout << -1;
      return 0;
    }
    ll tmp = (val-a[idx])/val2;
    ans += tmp;
    val -= val2*tmp;
    b[idx] = val;
    if(val > a[idx]) q.emplace(val,idx);
  }
  cout << ans;
  return 0;
}