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_imports)]
#![allow(dead_code)]

use proconio::input;
use std::collections::*;

fn prime_factorization(x: i64) -> std::collections::BTreeMap<i64, i64> {
    (std::iter::once(2).chain((3..).step_by(2)))
        .scan(x, |state, i| match *state {
            1 => None,
            _ => {
                let divisor = if i * i <= *state { i } else { *state };
                let mut index = 0;
                while *state % divisor == 0 {
                    *state /= divisor;
                    index += 1;
                }
                Some((divisor, index))
            }
        })
        .filter(|&(_, index)| index > 0)
        .collect()
}

fn main() {
    input! {
        n:i64
    }
    let ans: i64 = prime_factorization(n)
        .iter()
        .map(|(_, &index)| {
            (1..)
                .take_while(|&i| index >= i * (i + 1) / 2)
                .last()
                .unwrap()
        })
        .sum();
    println!("{}", ans);
}