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_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use im_rc::HashMap;
use text_io::*;

fn main() {
    let n: usize = read!();
    if n == 1 {
        println!("0");
        std::process::exit(0);
    } else {
        let vec = trial_division(n);
        let mut ans = 0;
        for (_key, num) in vec {
            ans += f(num);
        }
        println!("{}", ans);
    }
}

fn f(u: usize) -> usize {
    let mut u = u;
    if u == 1 {
        return 1;
    } else {
        for i in 1..u {
            if u - i <= i {
                return i;
            } else {
                u -= i;
            }
        }
    }
    10000
}

fn trial_division(mut n: usize) -> HashMap<usize, usize> {
    let mut primes = HashMap::new();
    let mut i = 2;

    //  n を root n 以下の整数で割り切れるまで割っていく
    while i * i <= n {
        while n % i == 0 {
            n /= i;
            if primes.contains_key(&i) {
                let mut x = primes.get_mut(&i).unwrap();
                *x += 1;
            } else {
                primes.insert(i, 1);
            }
        }
        i += 1;
    }

    // 最後にnが素数になっている場合はそれ自身も素因数に含めて終了
    if n > 1 {
        if primes.contains_key(&n) {
            let mut x = primes.get_mut(&n).unwrap();
            *x += 1;
        } else {
            primes.insert(n, 1);
        }
    }
    primes
}