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: Accepted)

use std::collections::HashMap;
use std::error::Error;
use std::io;

pub fn main() {
    if let Err(_e) = run() {
        panic!("failed.");
    }
}

fn run() -> Result<(), Box<dyn Error>> {
    let mut input = String::new();
    io::stdin().read_line(&mut input)?;
    let n: i64 = input.trim().parse()?;
    let map = resolve_prime(n);
    let mut result = 0;
    for &x in map.values() {
        let mut temp = x;
        let mut i = 1;
        while temp >= i {
            temp -= i;
            result += 1;
            i += 1;
        }
    }
    println!("{}", result);

    Ok(())
}

fn resolve_prime(mut n: i64) -> HashMap<i64, i64> {
    let mut i = 2i64;
    let mut result = HashMap::new();
    while i * i <= n {
        if i != 2 && i % 2 == 0 {
            i += 1;
            continue;
        }
        if n % i == 0 {
            let count = result.entry(i).or_insert(0);
            *count += 1;
            n /= i;
        } else {
            i += 1;
        }
    }
    if n != 1 {
        let count = result.entry(n).or_insert(0);
        *count += 1;
    }
    result
}