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
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

# check reachability B -> A ?
# 隣接和table
P = [B[(i-1)%N] + B[(i+1)%N] for i in range(N)]
D = deque(i for i in range(N) if B[i] - P[i] >= A[i])
ans = 0
nmatch = [B[i] == A[i] for i in range(N)].count(True)
while D:
    upd = False
    i = D.popleft()
    if B[i] == A[i]:
        nmatch += 1
        continue
    if B[i] <= P[i]:
        continue
    if B[i] - P[i] >= A[i]:
        n = (B[i] - A[i]) // P[i]
        P[(i-1)%N] -= P[i] * n
        P[(i+1)%N] -= P[i] * n
        B[i] -= P[i] * n
        upd = True
        ans += n
        if B[i] == A[i]:
            nmatch += 1
        if B[(i-1)%N] - P[(i-1)%N] >= A[(i-1)%N]:
            D.append((i-1)%N)
        if B[(i+1)%N] - P[(i+1)%N] >= A[(i+1)%N]:
            D.append((i+1)%N)
    if not upd:
        break

print(ans if nmatch == N else -1)
