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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
    $num = intval(trim(fgets(STDIN)));
    if($num ==1){
        echo "0\n";
        exit;
    }
    $standard = $num;
    
    $soinsu = array();
    for($j=2;$j<=sqrt($standard);$j++){
        $soinsu[$j] = 0;
    }
    start:
    for($i=2;$i<=sqrt($standard);$i++){
        if($num%$i ==0){
            // echo $num." ".$i."\n";
            $num = $num/$i;
            $soinsu[$i]++;
            goto start;
        }    
    }
    $count=0;
    $k=1;
    $h=2;
    $last_key = array_key_last((array)$soinsu);
    //素因数の個数を記録した配列に対して１，２，３，４，５と引く数を増やしていきマイナスになる
    for($h=2;$h<=$last_key;$h++){
        while($soinsu[$h]>0){
            // echo $soinsu[$h]."\n";
            $soinsu[$h] = $soinsu[$h] - $k;
            $k++;
            $count++;
        }
    }
    if($count==0){
        echo "1\n";
    }else{
        echo $count."\n";
    }
?>