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 PyPy3 (7.3.0) (judge verdict: Accepted)

#!/usr/bin/env python3
#%% for atcoder uniittest use
import sys
input= lambda: sys.stdin.readline().rstrip()
def pin(type=int):return map(type,input().split())
def tupin(t=int):return tuple(pin(t))
def lispin(t=int):return list(pin(t))
#%%code
def resolve():
    N,=pin()
    if N==1:
        print(0)
        return

        
    n=int(N**0.5)+1
    divs=dict()
    temp=N
    for i in range(2,n+4):
        while (1):
            if temp%i !=0:
                break
            divs.setdefault(i,0)
            divs[i]+=1
            temp/=i
    if temp>1:
        divs.setdefault(temp,1)
    #print(divs)
    check=1
    ans=0
    for p,q in divs.items():
        check*=p**q
        temp2=0
        for i in range(1,50):
            temp2+=i
            if temp2>q:
                break
            ans+=1
            
            
        #print(ans)
    print(ans)
    assert N== check
    
#%%submit!
resolve()