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];
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	scanf("%d",&a[i]);
	for (int i=0;i<n;i++)
	scanf("%d",&b[i]);
	int l=n-1;
	long long ans=0;
	for (int i=0;;i=(i+1)%n)
	{
		bool ch=0;
		int x=b[(i-1+n)%n],y=b[i],z=b[(i+1)%n];
		if (y>x+z && y!=a[i])
		{
			b[i]%=(x+z);
			if (b[i]<a[i])
			{
				if ((a[i]-b[i])%(x+z))
				{
					printf("-1");
					return 0;
				}
				b[i]=a[i];
			}
			ans+=(y-b[i])/(x+z);
			l=i;
			ch=1;
		}
		if (l==i && !ch)
		break;
	}
	for (int i=0;i<n;i++)
	{
		if (a[i]!=b[i])
		{
			printf("-1");
			return 0;
		}
	}
	printf("%lld",ans);
}