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 proconio::input;

fn primes() -> impl Iterator<Item = i64> {
    let mut multiples = HashMap::with_capacity(10000);
    let iter = (3..).step_by(2).filter_map(move |i| {
        let (prime_or_none, factor) = match multiples.remove(&i) {
            Some(f) => (None, f),
            None => (Some(i), i * 2),
        };

        (1..)
            .map(|j| i + j * factor)
            .skip_while(|m| multiples.contains_key(m))
            .next()
            .map(|m| multiples.insert(m, factor));

        prime_or_none // filter_map()は次の素数が見つかるまでリトライする。
    });
    // 最初に2を返してから、他の素数を返す。
    Box::new((2..).take(1).chain(iter))
}


fn main() {
    input!{
        mut n: i64,
    };
    let end = (n as f64).sqrt().ceil() as i64;
    let mut p = HashMap::new();
    for i in primes() {
        if i > end {
            break;
        }
        while n % i == 0 {
            *p.entry(i).or_insert(0) += 1;
            n /= i;
        }
        if n == 1 {
            break;
        }
    }
    if n != 1 {
        *p.entry(n).or_insert(0) += 1;
    }
    let mut ans = 0;
    for &m in p.values() {
        let mut m = m;
        for i in 1.. {
            m -= i;
            if m >= 0 {
                ans += 1;
            } else {
                break;
            }
        }
    }
    println!("{}", ans);
}