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 proconio::input;
use proconio::marker::{Chars, Usize1};
use std::iter::Iterator;
use std::collections::HashMap;
use std::cmp::*;
use itertools::Itertools;


//素因数分解
pub fn prime_factor(mut n: i64) -> HashMap<i64, i64> {
    let mut res: HashMap<i64, i64> = HashMap::new();
    let mut i = 2;
    while i * i <= n {
      while n % i == 0 {
        let v = match res.get(&i) {
          Some(v) => *v + 1,
          None => 1,
        };
        res.insert(i, v);
        n /= i;
      }
      i += 1;
    }
    if n != 1 {
      res.insert(n, 1);
    }
    res
  }

#[allow(unused_macros)]
macro_rules! max {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      max($x,max!($($xs)+))
    };
  }
  
  #[allow(unused_macros)]
  macro_rules! min {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      min($x,min!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn main() {
    input! {
        n: i64,
    }

    let s = prime_factor(n);

    let mut cnt = 0;
    for (k,v) in s {
        let mut vv = v;
        let mut b = 1;
        while b <= vv {
            vv -=b;
            b += 1;
            cnt += 1;
        }
    }

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