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<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=200005;
int n;
int a[N],b[N];
priority_queue<pair<int,int>,vector<pair<int,int> > >q;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		scanf("%d",&b[i]);
	for(int i=1;i<=n;i++)
		q.push(make_pair(b[i],i));
	a[0]=a[n],b[0]=b[n];
	a[n+1]=a[1],b[n+1]=b[1];
	long long ans=0;
	while(!q.empty())
	{
		int u=q.top().second;
		q.pop();
		if(a[u]==b[u]) continue;
		int x=b[u-1]+b[u+1];
		if(a[u]>b[u])
		{
			printf("-1");
			return 0;
		}
		if(b[u]-a[u]<x)
		{
			printf("-1");
			return 0;
		}
		int t=(b[u]-a[u])/x;
		ans+=t;
		b[u]-=t*x;
		if(u==1) b[n+1]=b[1];
		if(u==n) b[0]=b[n];
		if(a[u]!=b[u]) q.push(make_pair(b[u],u));
	}
	printf("%lld",ans);
	return 0;
}