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 n and 10 are coprimes, otherwise returns -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 

cycle(22) = -1 since 1/22 ~ 0.0 45 45 45 45 ...
```

Difficulty: EASY
Starter code: def cycle(n) :
	

Solutions:
import math

def cycle(n):
	if n % 2 == 0 or n % 5 == 0:
		return -1
	k = 1
	while pow(10, k, n) != 1:
		k += 1
	return k


def cycle(n):
	if not n % 2 or not n % 5:
		return -1
	(x, mods) = (1, set())
	while x not in mods:
		mods.add(x)
		x = 10 * x % n
	return len(mods)


def cycle(n):
	if n % 5 == 0 or n % 2 == 0:
		return -1
	(remainders, r) = ([1], 0)
	while r != 1:
		r = remainders[-1] * 10 % n
		remainders.append(r)
	return len(remainders) - 1


def cycle(n):
	if n % 2 and n % 5:
		for i in range(1, n):
			if pow(10, i, n) == 1:
				return i
	return -1


def cycle(n):
	if n % 2 == 0 or n % 5 == 0:
		return -1
	m = 1
	for p in range(1, n):
		m = 10 * m % n
		if m == 1:
			return p


def cycle(n):
	if not (n % 2 and n % 5):
		return -1
	return next(filter(lambda x: pow(10, x, n) == 1, range(1, n)))


def cycle(n):
	if not n % 2 or not n % 5:
		return -1
	(m, s) = (1, set())
	while m not in s:
		s.add(m)
		m = 10 * m % n
	return len(s)


from math import ceil, log10

def cycle(d):
	if d % 5 == 0 or d % 2 == 0:
		return -1
	n = 10 ** int(ceil(log10(d)))
	first = n % d
	n *= 10
	for i in range(99999999):
		n %= d
		if n == first:
			return i + 1
		n *= 10


def cycle(n):
	if n % 2 == 0 or n % 5 == 0:
		return -1
	c = len(str(n))
	m = 10 ** c % n
	while m != 1:
		c += 1
		m = m * 10 % n
	return c


def cycle(i):
	if i % 2 == 0 or i % 5 == 0:
		return -1
	gen = 10
	n = 1
	while gen % i != 1:
		gen = gen % i * 10
		n = n + 1
	return n