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

if any(a>b for a,b in zip(A,B)):
    print(-1)
    exit()

def pb(i):
    return i-1 if i-1 >= 0 else N-1
def nx(i):
    return i+1 if i+1 < N else 0

hq = []
for i,(a,b) in enumerate(zip(A,B)):
    if a <= b - B[pb(i)] - B[nx(i)]:
        hq.append((-b,i))
heapq.heapify(hq)

ans = 0
while hq:
    _,i = heapq.heappop(hq)
    if B[i] == A[i]: continue
    l = B[pb(i)]
    r = B[nx(i)]
    if B[i]-l-r < A[i]: continue
    k = (B[i]-A[i]) // (l+r)
    ans += k
    B[i] -= k*(l+r)
    ll = B[pb(pb(i))]
    rr = B[nx(nx(i))]
    if A[nx(i)] <= r - B[i] - rr:
        heapq.heappush(hq, (-r,nx(i)))
    if A[pb(i)] <= l - B[i] - ll:
        heapq.heappush(hq, (-l,pb(i)))

if all(a==b for a,b in zip(A,B)):
    print(ans)
else:
    print(-1)