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
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res

def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

#A
def A():
    n = I()

    return

#B
def B():
    n = I()

    return

#C
def C():
    n = I()
    a = LI()
    b = LI()
    q = []
    for i in range(n):
        if b[i] > a[i]:
            heappush(q,(-b[i],i))
    ans = 0
    while q:
        bi,i = heappop(q)
        r = (i+1)%n
        l = i-1
        s = b[l]+b[r]
        if a[i] < s:
            m = b[i]//s
            if m:
                ans += m
                b[i] %= s
                if b[i] != a[i]:
                    heappush(q,(-b[i],i))
        else:
            k = -bi-a[i]
            m = k//s
            b[i] -= m*s
            ans += m
            if b[i] != a[i]:
                break
    if a == b:
        print(ans)
    else:
        print(-1)
    return

#D
def D():

    return

#E
def E():
    n = I()

    return

#F
def F():
    n = I()

    return

#Solve
if __name__ == "__main__":
    C()
