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
def main():
	n=int(input())
	src = [int(item) for item in input().split()]
	dst = [int(item) for item in input().split()]
	ans=0
	def	SandQ(i):
		ri=i+1
		li=i-1
		if i==0:li=n-1
		if i==n-1:ri=0
		x=0
		if dst[i]-src[i]>=dst[ri]+dst[li]:
			x=(dst[i]-src[i])//(dst[ri]+dst[li])
			dst[i]-=(dst[ri]+dst[li])*x
			x+=SandQ(li)
			x+=SandQ(ri)
		return x;
	for i in range(n):
		ans+=SandQ(i)
	for i in range(n):
		if src[i]!=dst[i]:
			print(-1)
			exit()
	else:print(ans)
main()
