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)

#[allow(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($( { splits.next().unwrap().parse::<$t>().unwrap() },)*)
    }}
}

#[allow(dead_code)]
fn prime_factorization(n: u64) -> std::collections::HashMap<u64, u32> {
    let primes = primes((n as f32).sqrt().ceil() as u64);
    let mut h = std::collections::HashMap::new();
    let mut current = n;
    while current != 1 {
        let mut found = false;
        for p in &primes {
            if current % p == 0 {
                let count = h.entry(*p).or_insert(0);
                *count += 1;
                current /= *p;
                found = true;
            }
        }
        if !found {
            let count = h.entry(n).or_insert(0);
            *count += 1;
            break;
        }
    }
    return h;
}

#[allow(dead_code)]
fn primes(n: u64) -> Vec<u64> {
    if n < 3 {
        return vec![];
    }
    let mut res: Vec<u64> = vec![2];
    for i in 3..n {
        let i = i as u64;
        let root = ((n as f64).sqrt().ceil() as usize + 1) as u64;
        let composite = res.iter().take_while(|&&x| x <= root).any(|x| i % x == 0);
        if !composite {
            res.push(i);
        }
    }
    return res;
}

#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (n,) = input!(u64);

    let factors = prime_factorization(n);

    let mut total = 0;

    for (k, v) in factors {
        let mut count = 1;
        let mut vcount = v;
        while vcount >= count {
            vcount -= count;
            count += 1;
        }
        total += count - 1;
    }

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

fn main() {
    solve();
}