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,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
using P = pair<ll,ll>;

int main(){
  ll n, l, m, r, s, c, d = 0;
  cin >> n;
  ll a[n], b[n];
  priority_queue<P> que;
  rep(i,n) cin >> a[i];
  rep(i,n) {
    cin >> b[i];
    que.push(P(b[i],i));
  }
  while(!que.empty()) {
    P p = que.top(); que.pop();
    m = p.second;  l = (m-1+n) % n, r = (m+1) % n;
    s = b[l] + b[r]; c = min(b[m]/s,(b[m]-a[m])/s);
    if(c > 0) {
      b[m] -= s*c; que.push(P(b[m],m)); d += c;
    } 
  }
  rep(i,n) {
    if(a[i] != b[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << d << endl;
  return 0;
}