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 collections import deque

def resolve():
    n = int(input())
    aa = list(map(int, input().split()))
    bb = list(map(int, input().split()))

    q = deque()
    diffs = set()

    for i in range(n):
        if bb[i] != aa[i]:
            diffs.add(i)

        bef_i = i-1 if i > 0 else n-1
        aft_i = i+1 if i < n-1 else 0
        if bb[i] > aa[i] and bb[i] > bb[bef_i] + bb[aft_i]:
            q.append(i)

    count = 0

    while len(q) > 0:
        i = q.popleft()

        bef_i = i-1 if i > 0 else n-1
        bef2_i = bef_i-1 if bef_i > 0 else n-1
        aft_i = i+1 if i < n-1 else 0
        aft2_i = aft_i+1 if aft_i < n-1 else 0

        count += (bb[i] - aa[i]) // (bb[bef_i] + bb[aft_i])
        bb[i] = (bb[i] - aa[i]) % (bb[bef_i] + bb[aft_i]) + aa[i]

        if bb[i] == aa[i]:
            diffs.remove(i)

        if bb[bef_i] > aa[bef_i] and bb[bef_i] > bb[bef2_i] + bb[i]:
            q.append(bef_i)

        if bb[aft_i] > aa[aft_i] and bb[aft_i] > bb[aft2_i] + bb[i]:
            q.append(aft_i)

    if len(diffs) > 0:
        print(-1)
        return

    print(count)
resolve()