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 Rust (1.42.0) (judge verdict: Wrong Answer)

use std::io;

fn main() {
    let mut S = String::new();
    io::stdin().read_line(&mut S);
    let nf :f64 = S.trim().parse().unwrap();
    let mut n :usize = nf as usize;
    let mut count :usize = 0;
    let mut i :usize = 1;
    if n == 1{
        println!("0");
        return ();
    }
    loop{
        if n % usize::pow(2,i as u32) == 0{
            count += 1;
            n = n/usize::pow(2,i as u32);
        }
        if usize::pow(2,i as u32) > n{
            break;
        }
        i += 1;
    }
    for j in (3..=nf.sqrt() as usize).step_by(2) {
        if n % j == 0{
                n = n/i;
                count = count + 1;
        }
        if n < j {
            break;
        }
    }
    if count == 0 {
        count = 1;
    }
    println!("{}",count);
}