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 pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=200010;
const ll mod=1e9+7;


int main(){
  int n;
  cin >> n;
  vector<ll> a(n),b(n);
  rep(i,n){ cin >> a[i]; }
  priority_queue<pair<ll,int>> q;
  rep(i,n){ cin >> b[i]; q.push({b[i],i}); }
  ll cnt = 0;
  while(!q.empty()){
    ll p; int pos;
    tie(p,pos) = q.top(); q.pop();
    int posl = pos-1; if(posl<0) posl+=n;
    int posr = pos+1; if(posr>=n) posr-=n;
    ll tmp = (b[pos]-a[pos])/(b[posl]+b[posr]);
    b[pos] -= (b[posl]+b[posr])*tmp;
    cnt+=tmp;
    if(a[pos]!=b[pos] && tmp){ q.push({b[pos],pos}); }
  }
  if(a==b){
    cout << cnt << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}