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 heapify, heappush, heappop

def max2(x, y): return x if x >= y else y

def solve(As, Bs):
    for A, B in zip(As, Bs):
        if A > B:
            return -1

    PQ = [(-B, i) for i, (A, B) in enumerate(zip(As, Bs)) if A != B]
    heapify(PQ)
    ans = 0
    while PQ:
        B, i = heappop(PQ)
        A, B = As[i], -B
        sumNeib = Bs[i-1] + Bs[(i+1)%N]

        if B <= sumNeib or B-sumNeib < A:
            return -1
        num = (B - max2(A, sumNeib)) // sumNeib
        B -= sumNeib * num
        if B-sumNeib >= A:
            B -= sumNeib
            num += 1

        if A != B:
            heappush(PQ, (-B, i))
        Bs[i] = B
        ans += num

    return ans


N = int(input())
As = list(map(int, input().split()))
Bs = list(map(int, input().split()))

ans = solve(As, Bs)
print(ans)
