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[200005],b[200005],i,n;
long long ans;
struct str{
	int x,s;
};
bool operator <(str a,str b)
{
	return a.s<b.s;
}
priority_queue<str> q;
int main()
{
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	for(i=0;i<n;i++)
	{
		scanf("%d",&b[i]);
		q.push((str){i,b[i]});
	}
	while(!q.empty())
	{
		str x=q.top();
		q.pop();
		if(a[x.x]==b[x.x])
			continue;
		int s=b[(x.x-1+n)%n]+b[(x.x+1)%n];
		if(b[x.x]<=s)
			break;
		if(a[x.x]>=s)
		{
			if((b[x.x]-a[x.x])%s==0)
			{
				ans+=(b[x.x]-a[x.x])/s;
				b[x.x]=a[x.x];
			}
			else
				break;
		}
		else
		{
			ans+=b[x.x]/s;
			b[x.x]%=s;
			q.push((str){x.x,b[x.x]});
		}
	}
	for(i=1;i<=n;i++)
		if(a[i]!=b[i])
		{
			printf("-1");
			return 0;
		}
	cout<<ans;
}
