Codewars - 1/n Cycle

'''
Let be n an integer prime with 10 e.g. 7.
1/7 = 0.142857 142857 142857 ....
We see that the decimal part has a cycle: 142857. The length of this cycle is 6. In the same way:
1/11 = 0.09 09 09 .... Cycle length is 2.
Task
Given an integer n (n > 1) the function cycle(n) returns the length of the cycle if there is one otherwise (n and 10 not coprimes) return -1.
Examples
cycle(5) = -1
cycle(13) = 6 -> 0.076923 076923 0769
cycle(21) = 6 -> 0.047619 047619 0476
cycle(27) = 3 -> 0.037 037 037 037 0370
cycle(33) = 2 -> 0.03 03 03 03 03 03 03 03
cycle(37) = 3 -> 0.027 027 027 027 027 0
cycle(94) = -1
'''

def cycle(n):
    if n % 10 == 0:
        print("There is a problem while using function " + cycle.__name__)
        return -1
    klasma = str(1 / n)
    # I will take the quantity after "0.", so I will start from index 2
    quantity = klasma[2:]
    print("n = " + str(n) + " ----> quantity = " + str(quantity))
    N = len(quantity)
    # First, I have to check if the 1st digit appears again
    # If this case doesn't happen, then we know that there is no posiibility of a cycle and we return  -1
    first = quantity[0]
    appears = False
    index = -1000
    for i in range(1, N):
        if quantity[i] == first:
            appears = True
            index = i
            break
    if appears == False:
        return -1
    # We continue here since we have found that the first digit appears again
    # We also know that Python "holds" max = 17 decimal places
    flag = True
    for i in range(0, N - index - 1):       # I put "-1" for precision errors
        for j in range(0, index):
            if quantity[i] != quantity[i+index]:
                # print(quantity[i], quantity[i+index])
                flag = False
    if flag == False:
        return -1
    return index


# MAIN FUNCTION
for n in range(2, 100):
    print(cycle(n))
    print()