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: Accepted)

import math
import sys

#素数か(x>=2)
def prime_check(x): #x>=2
    if x==2:
        return True,2
    elif x%2==0:
        return False,2
    else:
        limit=int(math.sqrt(x))
        for factor in range(3,limit+1,2):
            if x%factor==0:
                return False,factor
        return True,x

#素因数分解(x>=2)
def prime_factorization(x): #x>=2
    factor_dictionary={}
    factor_set=set()
    judge,factor=prime_check(x)
    while True:
        if factor in factor_set:
            factor_dictionary[factor]+=1
        else:
            factor_dictionary[factor]=1
            factor_set.add(factor)

        if judge:
            return factor_dictionary,factor_set
            break
        else:
            x//=factor
            judge,factor=prime_check(x)

n=int(input())
Fd,Fs=prime_factorization(n)

def culc(x):
    temp=1
    while True:
        if x>=temp:
            x-=temp
            temp+=1
        else:
            return temp-1
            break

if n==1:
    print(0)
    sys.exit()

ans=0
while Fs:
    now=Fs.pop()
    ans+=culc(Fd[now])

print(ans)