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 MOD 1000000007
#define BIG 1000000010
#define EPS 1e-9
#define fst first
#define scd second

#define debug(x) cout<<x<<endl;
#define repi(i,x,n) for(int i=x;i<n;i++)
#define rep(i,n) repi(i,0,n)
#define repn(i,n) for(int i=n;i>=0;i--)
#define int long long
#define endl "\n"


signed main(){
  cin.tie(0);	
  ios::sync_with_stdio(false);
  int a[200000],b[200000];
  int n;
  cin>>n;
  rep(i,n) cin>>a[i];
  rep(i,n) cin>>b[i];
  bool flag=true;
  int ans=0;
  while(flag){
    flag=false;
    rep(i,n){
      if(b[(i+n-1)%n]+b[(i+1)%n] <= b[i]-a[i]){
	ans+=(b[i]-a[i])/(b[(i-1+n)%n]+b[(i+1)%n]);
	b[i]=((b[i]-a[i])%(b[(i-1+n)%n]+b[(i+1)%n]))+a[i];
	
	flag=true;
      }
    }
  }
  rep(i,n){
    if(a[i]!=b[i]){cout<<-1<<endl;return 0;}
  }
  //cout<<-1 % 3 <<endl;
  cout<<ans<<endl;
  return 0;
}