Problem:
Score : 400 points

Problem Statement

Given is a positive integer N. Consider repeatedly applying the operation below on N:

First, choose a positive integer z satisfying all of the conditions below:

z can be represented as z=p^e, where p is a prime number and e is a positive integer;

z divides N;

z is different from all integers chosen in previous operations.

Then, replace N with N/z.

Find the maximum number of times the operation can be applied.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum number of times the operation can be applied.

Sample Input 1

24

Sample Output 1

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=12.)

Choose z=3 (=3^1). (Now we have N=4.)

Choose z=4 (=2^2). (Now we have N=1.)

Sample Input 2

1

Sample Output 2

0

We cannot apply the operation at all.

Sample Input 3

64

Sample Output 3

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=32.)

Choose z=4 (=2^2). (Now we have N=8.)

Choose z=8 (=2^3). (Now we have N=1.)

Sample Input 4

1000000007

Sample Output 4

1

We can apply the operation once by, for example, making the following choice:

z=1000000007 (=1000000007^1). (Now we have N=1.)

Sample Input 5

997764507000

Sample Output 5

7

Solution in Python (3.8.2) (judge verdict: Runtime Error)

import sympy

primes= list(sympy.primerange(1, 10**6))

n=int(input())

prime_factors= {}

while n != 1:
	for p in primes:
		if n%p==0:
			if p in prime_factors:
				prime_factors[p]+=1
			else:
				prime_factors[p]=1
			n/= p
			break
		elif p==999983:
			prime_factors[n]=1
			n=1
			break
tot= 0
for p in prime_factors:
	if prime_factors[p]>=36:
		tot+= 8
	elif prime_factors[p]>=28:
		tot+=7
	elif prime_factors[p]>=21:
		tot+=6
	elif prime_factors[p]>=15:
		tot+=5
	elif prime_factors[p]>=10:
		tot+=4
	elif prime_factors[p]>=6:
		tot+=3
	elif prime_factors[p]>=3:
		tot+=2
	else:
		tot+=1


print(tot)