Given an integer `n` return `"odd"` if the number of its divisors is odd. Otherwise return `"even"`.

**Note**: big inputs will be tested.

## Examples:

All prime numbers have exactly two divisors (hence `"even"`).

For `n = 12` the divisors are `[1, 2, 3, 4, 6, 12]` – `"even"`.

For `n = 4` the divisors are `[1, 2, 4]` – `"odd"`.

Difficulty: introductory

Starter code:

def oddity(n):
	

Solutions:
def oddity(n):
    #your code here
    return 'odd' if n**0.5 == int(n**0.5) else 'even'

oddity=lambda n: ["odd","even"][n**.5%1!=0]

import math
def oddity(n):
    return math.sqrt(n) % 1 == 0 and 'odd' or 'even'

import math
def oddity(n):
    
    return "even" if math.sqrt(n)%1 else "odd"


import math
def oddity(n):
    if int(math.sqrt(n))**2 == n:
        return 'odd'
    return 'even'

def oddity(n):
    counter = 0
    for x in range(1, int(n**.5)+1):
        a,b = divmod(n,x)
        if not b:
            if a!=x: counter += 2
            else: counter += 1
    return ('even', 'odd')[counter%2]

def oddity(n):
    return 'odd' if (n ** 0.5).is_integer() else 'even'

def oddity(n):
    return 'even' if n ** 0.5 % 1 else 'odd'

def oddity(n):
    return 'odd' if int(n**0.5)**2 == n else 'even'

def oddity(n):
    return 'odd' if round(n ** 0.5) ** 2 == n else 'even'