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
n = int(input())
b = list(map(int,input().split()))
a = list(map(int,input().split()))
lock = [0]*n

ans = 0
top = []
for i in range(n):
  if a[i] < b[i]:
    print(-1)
    exit()
  if a[i] == b[i]:
    lock[i] = 1
    continue
  if a[i-1]+a[(i+1)%n]<a[i]:
    top.append(i)
while True:
  while top:
    x = top.pop()
    ne = a[x-1]+a[(x+1)%n]
    if (a[x]-b[x])%ne == 0:
      ans += (a[x]-b[x])//ne
      lock[x] = 1
      a[x] = b[x]
    else:
      ans += a[x]//ne
      a[x] %= ne
    for i in (x+1,x-1):
      if a[i-1]+a[(i+1)%n]<a[i%n]:
        top.append(i%n)
  if lock.count(1) == n:
    print(ans)
    exit()
  elif not top:
    print(-1)
    exit()