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 heapq import heappush, heappop

N = int(input())
A = [int(i) for i in input().split()]
B = [int(i) for i in input().split()]

def solve() :
    h = []
    for i in range(N) :
        if A[i] == B[i] :
            continue
        heappush(h, (-B[i], i))

    ret = 0
    while h :
        _, i = heappop(h)
       
        m, M = B[(i-1)%N], B[(i+1)%N]
        if m > M :
            m, M = M, m
        
        if M >= A[i] :
            x = (B[i] + m - 1) // (m + M)
            B[i] -= x * (m + M)
            ret += x
        else :
            if (B[i] - A[i]) % (m + M) != 0 :
                return -1
            ret += (B[i] - A[i]) // (m + M)
            B[i] = A[i]
        
        if A[i] > B[i] :
            return -1
            
        elif A[i] < B[i] :
            heappush(h, (-B[i], i))
    
    return ret
    
print(solve())