Create a function named `divisors`/`Divisors` that takes an integer `n > 1` and returns an array with all of the integer's divisors(except for 1 and the number itself), from smallest to largest. If the number is prime return the string '(integer) is prime' (`null` in C#) (use `Either String a` in Haskell and `Result, String>` in Rust).

#### Example:

```python
divisors(12); #should return [2,3,4,6]
divisors(25); #should return [5]
divisors(13); #should return "13 is prime"
```

Source: codewars | Difficulty: EASY
Tags: Mathematics

Starter code: def divisors(integer):
	

Solutions:
def divisors(num):
	l = [a for a in range(2, num) if num % a == 0]
	if len(l) == 0:
		return str(num) + ' is prime'
	return l


def divisors(integer):
	a = []
	for i in range(2, integer):
		if integer % i == 0:
			a.append(i)
	return a if a else str(integer) + ' is prime'


def divisors(integer):
	return [n for n in range(2, integer) if integer % n == 0] or '{} is prime'.format(integer)


import math

def divisors(n):
	o = [i for i in range(2, int(math.ceil(n / 2) + 1)) if n % i == 0]
	return o if len(o) > 0 else '%d is prime' % n


def divisors(integer):
	lst = []
	for i in range(2, integer):
		if integer % i == 0:
			lst.append(i)
	if len(lst) == 0:
		return '%s is prime' % integer
	else:
		return lst


def divisors(n):
	divs = set()
	for t in range(2, int(n ** 0.5) + 1):
		(div, mod) = divmod(n, t)
		if mod == 0:
			divs.add(t)
			divs.add(div)
	return '{:d} is prime'.format(n) if len(divs) == 0 else sorted(list(divs))


def divisors(n):
	result = [i for i in range(2, n) if n % i == 0]
	return result if result else str(n) + ' is prime'


from numpy import arange

def divisors(integer):
	potential_divisors = arange(2, integer // 2 + 1)
	divisors = potential_divisors[integer % potential_divisors == 0]
	if len(divisors):
		return list(divisors)
	else:
		return '{integer} is prime'.format(integer=integer)