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;

macro_rules! get_one {
    ($t:ty) => {{
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse::<$t>().unwrap()
    }};

    ($t:ty; $n:expr) => {
        (0..$n).map(|_| get_one!($t)).collect::<Vec<_>>()
    };
}

fn main() {
    let n = get_one!(usize);

    let hm = factor(n as i64);

    let a: i64 = hm.iter()
        .map(|(_, &v)| {
            max_triangle_num(v)
        }).sum();
    println!("{}", a)
}

fn factor(n: i64) -> HashMap<i64, i64> {
    let mut n = n;
    let mut hm = HashMap::new();
    if n == 1 {return hm};

    let mut p = 2;
    while n != 1 && n >= p * p {
        while n % p == 0 {
            n /= p;
            let count = hm.entry(p).or_insert(0);
            *count += 1;
        }
        p += 1;
    }
    if n != 1 {
        hm.insert(n, 1);
    }
    hm
}

fn max_triangle_num (n: i64) -> i64 {
    let mut m = 1;
    while m * (m + 1) / 2 <= n {
        m += 1;
    }
    return m - 1
}

#[test]
fn test_max_triangle_num() {
    assert_eq!(max_triangle_num(1), 1);
    assert_eq!(max_triangle_num(2), 1);
    assert_eq!(max_triangle_num(3), 2);
    assert_eq!(max_triangle_num(4), 2);
    assert_eq!(max_triangle_num(5), 2);
    assert_eq!(max_triangle_num(6), 3);
}