Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3
import sys

from functools import reduce

import math

def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        if n % i == 0:
            return False
    return True

def sieve(n):
    n=int(n)
    # Create a boolean array "prime[0..n]" and initialize 
    # all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n + 1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * 2, n + 1, p): 
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False

    return [i for i, p in enumerate(prime) if p]

def inthroot(x,n):
    "calculate floor(x**(1/n))"
    return math.floor(x**(1/n))

def is_a_power(n):
    "return (a,b) if n=a**b otherwise throw ValueError"
    for b in sieve( math.log2(n) +1 ):
        if b == 0:
            raise ValueError
            
        a = inthroot(n,b)
        if a**b == n:
            return a,b
    raise ValueError
    


def smooth_factorization(n):
    "return (p,e) where p is prime and n = p**e if such value exists, otherwise throw ValueError"
    e=1
    p=n
    while True:
        try:
            p,n = is_a_power(p)
            e = e*n
        except ValueError:
            break
    if is_prime(p): 
        return p,e


# def factors(n):
#     return set(
#         reduce(
#             list.__add__,
#             ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0),
#         )
#     )

def factors(n):    
        l1, l2 = [], []
        for i in range(1, int(n ** 0.5) + 1):
            q,r = n//i, n%i     # Alter: divmod() fn can be used.
            if r == 0:
                l1.append(i) 
                l2.append(q)    # q's obtained are decreasing.
        if l1[-1] == l2[-1]:    # To avoid duplication of the possible factor sqrt(n)
            l1.pop()
        l2.reverse()
        return l1 + l2
        
def solve(N: int):
    count = 0
    picked_z = set([])

    while True:
        zs = list(factors(N))
        zs.remove(1)

        for i, z in enumerate(zs):
            is_prime_power = smooth_factorization(z)
            if is_prime_power and z not in picked_z:
                picked_z.add(z)
                N = N / z
                count += 1
                break

        if len(zs) == 0:
            break

    print(count)
    return


# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word

    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    solve(N)


if __name__ == "__main__":
    main()


Fixed submission:
#!/usr/bin/env python3
import sys

from functools import reduce

import math

def is_prime(n):
    if n % 2 == 0 and n > 2: 
        return False
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        if n % i == 0:
            return False
    return True

def sieve(n):
    n=int(n)
    # Create a boolean array "prime[0..n]" and initialize 
    # all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n + 1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * 2, n + 1, p): 
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False

    return [i for i, p in enumerate(prime) if p]

def inthroot(x,n):
    "calculate floor(x**(1/n))"
    return math.floor(x**(1/n))

def is_a_power(n):
    "return (a,b) if n=a**b otherwise throw ValueError"
    for b in sieve( math.log2(n) +1 ):
        if b == 0:
            raise ValueError
            
        a = inthroot(n,b)
        if a**b == n:
            return a,b
    raise ValueError
    


def smooth_factorization(n):
    "return (p,e) where p is prime and n = p**e if such value exists, otherwise throw ValueError"
    e=1
    p=n
    while True:
        try:
            p,n = is_a_power(p)
            e = e*n
        except ValueError:
            break
    if is_prime(p): 
        return p,e


def factors(n):
    return set(
        reduce(
            list.__add__,
            ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0),
        )
    )
        
def solve(N: int):
    count = 0
    picked_z = set([])

    while True:
        zs = list(factors(N))
        zs.remove(1)

        for _ in range(len(zs)):
                    
            z = min(zs)
            i = zs.index(z)

            is_prime_power = smooth_factorization(z)
            if is_prime_power and z not in picked_z:
                picked_z.add(z)
                N = N / z
                count += 1
                del zs[i]
                break

            del zs[i]

        if len(zs) == 0:
            break

    print(count)
    return


# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word

    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    solve(N)


if __name__ == "__main__":
    main()