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
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)

from heapq import heappop, heappush

N = int(input())
A = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]

q = []
push = lambda x,i: heappush(q, (-x<<30)+i)
for i,x in enumerate(B):
    push(x,i)

answer = 0
mask = (1<<30)-1
while q:
    i = heappop(q)&mask
    x = B[i]
    dx = x-A[i]
    if dx < 0:
        answer = -1
        break
    if dx == 0:
        continue
    left = i-1 if i >= 1 else N-1
    right = i+1 if i < N-1 else 0
    nbd = B[left] + B[right]
    t = dx // nbd
    if t == 0:
        answer = -1
        break
    B[i] -= nbd * t
    answer += t
    push(B[i],i)

print(answer)