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 998244353
priority_queue<pair<int,int> >q;
int a[200001];
int b[200001];
signed main()
{
	int n,mx=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]),mx=max(mx,a[i]);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		if(b[i]>a[i])
			q.push(make_pair(b[i],i));
		else if(b[i]<a[i])
		{
			printf("-1");
			return 0;
		}
	}
	long long ans=0;
	while(!q.empty())
	{
		int at=q.top().second;
		q.pop();
		int tot=((at==1)?b[n]:b[at-1])+((at==n)?b[1]:b[at+1]);
		int time=(b[at]-a[at])/tot;
		if(time==0)
		{
			printf("-1");
			return 0;
		}
		ans+=time;
		b[at]-=time*tot;
		if(b[at]!=a[at])
			q.push(make_pair(b[at],at));
	}
	printf("%lld",ans);
	return 0;
}
