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
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())


from heapq import heappush, heappop

n = ni()
a = list(li())
b = list(li())

# n = 10**5
# a = [1]*(10**5)
# b = [5] + [1]*(10**5-2) + [7]

que = []

for i, bi in enumerate(b):
    heappush(que, (-bi, i))

ans = 0
while que:

    cur, position = heappop(que)
    cur = -cur

    if cur - b[(position-1)%n] - b[(position+1)%n] < a[position]:
        continue

    ope = (cur - a[position]) // (b[(position-1)%n] + b[(position+1)%n])

    ans += ope
    b[position] -= ope * (b[(position-1)%n] + b[(position+1)%n])
    heappush(que, (-b[position], position))

ok = True
for ai, bi in zip(a, b):
    if ai != bi:
        ok = False

if ok:
    print(ans)
else:
    print(-1)