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::cmp::*;
use std::collections::*;
use std::io::*;
use std::str::*;

use std::string::*;

struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> Self {
        Self(r, std::io::BufWriter::new(w))
    }

    pub fn write<S: ToString>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.to_string().as_bytes()).unwrap();
    }

    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let s: String = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect();

        return s.parse::<T>().ok().unwrap();
    }

    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}

fn prime_factorization(n: i64) -> HashMap<i64, i64> {
    let mut n = n;
    let mut res = HashMap::new();
    let mut i = 2;
    while i * i <= n {
        while n % i == 0 {
            n /= i;
            let count = res.entry(i).or_insert(0);
            *count += 1;
        }
        i += 1;
    }

    if n > 1 {
        res.insert(n, 1);
    }

    return res;
}

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());
    let n = sc.read();
    let mut res = prime_factorization(n);
    let mut ans = 0;
    for a in res {
        let mut cnt = 0;
        let mut plus = 1;
        while cnt + plus <= a.1 {
            cnt += plus;
            plus += 1;
            ans += 1;
        }
    }

    println!("{}", ans);
}