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
words = lambda t : list(map(t, input().split()))



n = int(input())
a = words(int)
b = words(int)

def getNext(i):
    if i == n-1:
        return 0
    else:
        return i+1

def getPrev(i):
    if i == 0:
        return n-1
    else:
        return i-1

from collections import deque
q = deque()

def verify(i):
    if b[i] != a[i] and b[i] - (b[getNext(i)] + b[getPrev(i)]) >= a[i]:
        return True
    else:
        return False

for i in range(len(b)):
    if b[i] >= a[i] and verify(i):
        q.append(i)

ans = 0
succeed = True
while not len(q) == 0:
    i = q.popleft()
    ni = getNext(i)
    pi = getPrev(i)
    #print(i, b)
    d = b[ni] + b[pi]
    if b[i] % d == a[i] % d:
        ans += b[i] // d - (a[i] // d)
        b[i] = a[i]
    else:
        ans += b[i] // d
        b[i] %= d
    if b[i] < a[i]:
        succeed = False
        break
    if verify(ni):
        q.append(ni)
    if verify(pi):
        q.append(pi)

for i in range(len(b)):
    if a[i] != b[i]:
        succeed = False
        break

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