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::io;
use std::str::FromStr;
use std::collections::HashMap;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

pub fn prime_factors(number: u128) -> HashMap<u128, u128> {
    let limit = number;
    let mut number = number; 
    let mut prime_numbers: HashMap<u128, u128> = HashMap::new();
    let mut candidate = 2;

    while number > 1 {
        if candidate * candidate > limit {
            prime_numbers.insert(limit, 1);
            return prime_numbers;
        }
        while number % candidate == 0 {
            let n = prime_numbers.entry(candidate).or_insert(0);
            *n += 1;
            number /= candidate;
        }
        candidate += 1;
    }
    return prime_numbers;
}

fn main() {
    let mut n: u128 = input();
    let prime_nums = prime_factors(n);

    // println!("{:?}", prime_nums);

    let mut ans = 0;
    for c in prime_nums.values() {
        let mut c_buf = *c;
        let mut i = 1;
        let mut count = 0;
        while c_buf >= i {
            c_buf -= i;
            i += 1;
            count += 1;
        }
        ans += count;
    }
    println!("{}", ans);
}