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: Wrong Answer)

use proconio::fastout;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
use std::collections::HashMap;

#[fastout]
fn main() {
    input! {
        n: u64,
    }
    if n == 1 {
        println!("{}", 0);
        return;
    }

    let mut n_calc = n;
    let mut factor = HashMap::new();
    let mut f = 2;

    loop {
        if n_calc % f == 0 {
            n_calc /= f;
            let e = factor.entry(f).or_insert(0);
            *e += 1;
            continue;
        }
        f += 1;
        if f.pow(2) > n {
            break;
        }
    }
    let ans = if factor.is_empty() {
        1
    } else {
        let mut cnt = 0;
        for fi in factor.values_mut() {
            let mut i = 1;
            while *fi >= i {
                *fi -= i;
                i += 1;
                cnt += 1;
            }
        }
        cnt
    };
    println!("{}", ans);
}