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()))

que=[(-B[i],i) for i in range(N)]
heapq.heapify(que)
count=0
while que:
    val,index=heapq.heappop(que)
    if val>0:
        break
    if index!=N-1:
        a,b,c=B[index-1],B[index],B[index+1]
        if a==A[index-1] and c==A[index+1]:
            if B[index]%(a+c)==A[index]%(a+c):
                count+=(B[index]-A[index])//(a+c)
                B[index]=A[index]
            else:
                count=-1
                break
        else:
            if B[index]>a+c:
                count+=B[index]//(a+c)
                B[index]%=a+c
                heapq.heappush(que,(-B[index],index))
    else:
        a,b,c=B[index-1],B[index],B[0]
        if a==A[index-1] and c==A[0]:
            if B[index]%(a+c)==A[index]%(a+c):
                count+=(B[index]-A[index])//(a+c)
                B[index]=A[index]
            else:
                count=-1
                break
        else:
            if B[index]>a+c:
                count+=B[index]//(a+c)
                B[index]%=a+c
                heapq.heappush(que,(-B[index],index))

if A==B:
    print(count)
else:
    print(-1)
