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())
A=[int(i) for i in input().split()]
B=[int(i) for i in input().split()]
ans=0
while(True):
    flag=0
    #print(B)
    for i in range(N):
        tmp=B[(i-1)%N]+B[(i+1)%N]
        if B[i]>tmp and B[i]>A[i]:
            beforeans=ans
            if B[i]%tmp>=A[i]:
                ans+=B[i]//tmp
                B[i]%=tmp
            else:
                k=((A[i]-1-(B[i]%tmp))//tmp)+1
                ans+=B[i]//tmp
                B[i]%=tmp
                #print(i,1,B)
                ans-=k
                B[i]+=k*tmp
                #print(i,2,B)
            if ans!=beforeans:
                flag=1
    if flag==0:
        break
flag2=1
for i in range(N):
    if A[i]!=B[i]:
        flag2=0
        break
if flag2:
    print(ans)
else:
    print(-1)
