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 heapq

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

C = [(-b, i) for i, b in enumerate(B) if b > A[i]]
heapq.heapify(C)

count = 0
while C:
    c, i = heapq.heappop(C)
    c *= -1
    if A[i] == c:
        continue
    if i == N-1:
        x = B[N-2] + B[0]
    else:
        x = B[i-1] + B[i+1]
    if c <= x or c < A[i] or c-A[i] < x:
        count = -1
        break
    temp = (c-A[i]) // x
    c -= temp * x
    count += temp
    B[i] = c
    if c > A[i]:
        heapq.heappush(C, (-c, i))
print(count)