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
from heapq import heappush, heappop
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ans = 0
q = []
for i, x in enumerate(b):
	l, r = b[(i-1)%n], b[(i+1)%n]
	if l + r < x:
		heappush(q, (-x, i))

ng = False
while q:
	x, i = -q[0][0], q[0][1]
	#print(x, i)
	y = a[i]
	if y > x:
		ng = True
		break
	elif y == x:
		heappop(q)
		continue
	else:
		l, r = b[(i-1)%n], b[(i+1)%n]
		quo = (x-1)//(l+r)
		mod = x - quo*(l+r)
		if y >= mod:
			if (x-y) % (l+r) != 0:
				ng = True
				break
			else:
				ans += (x-y) // (l+r)
				b[i] = y
		else:
			ans += quo
			b[i] = mod
		heappop(q)
		if b[(i-2)%n] + b[i] < l:
			heappush(q, (-l, (i-1)%n))
		if b[(i+2)%n] + b[i] < r:
			heappush(q, (-r, (i+1)%n))
		#print(b[i])
			
if ng or a != b:
	print(-1)
else:
	print(ans)