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;

int a[200001],b[200001];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<n;i++)
		cin>>b[i];
	priority_queue<pair<int,int> > pq;
	long long int ans=0;
	for(int i=0;i<n;i++)
		if(a[i]!=b[i])
			pq.push(make_pair(b[i],i));
	while(!pq.empty()){
		auto now=pq.top();pq.pop();
		int neighbor=b[(now.second+1)%n]+b[(now.second+n-1)%n];
		if(b[now.second]-a[now.second]<neighbor){
			cout<<-1<<endl;
			return 0;
		}
		ans+=(b[now.second]-a[now.second])/neighbor;
		b[now.second]=a[now.second]+(b[now.second]-a[now.second])%neighbor;
		if(a[now.second]!=b[now.second])
			pq.push(make_pair(b[now.second],now.second));
	}
	cout<<ans<<endl;
}
