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::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("faild to read char") as char)
        .take_while(|c| !c.is_whitespace())
        .collect();
    token
        .parse()
        .ok()
        .expect("faild to parse token")
}


fn main(){
    let n: u64 = read();
    let mut m = n;
    let mut ret = 0;

    for i in 2..n {
        if n < i * i {
            break;
        }

        let mut sum = 0;
        while m % i == 0 {
            sum += 1;
            m /= i;
        }

        if sum == 1 || sum == 2 {
            ret += 1;
            continue;
        }

        for j in 1..=sum {
            if sum < j * (j + 1) / 2 {
                ret += j - 1;
                //println!("{} {}", sum, i);
                break;
            }
        }
    }

    if m > 1 {
        ret += 1;
    }
    println!("{}", ret);
}