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 text_io::read;
use whiteread::{parse_line, reader::Result};

fn primes(max: &i64) -> Vec<i64> {
    let mut primes: Vec<i64> = Vec::new();
    let mut isprime: bool;

    for x in 2..*max {
        if x.pow(2) > *max {
            break;
        }

        isprime = true;
        for y in &primes {
            if y.pow(2) > x {
                break;
            }

            if x % y == 0 {
                isprime = false;
                break;
            }
        }

        if isprime {
            // println!("{} is prime", x);
            primes.push(x)
        }
    }

    isprime = true;
    for y in &primes {
        if *max % y == 0 {
            isprime = false;
            break;
        }
    }

    if isprime {
        // println!("{} is prime", x);
        primes.push(*max);
    }

    primes
}

fn main() -> Result<()> {
    let n: i64 = read!();

    if n == 1 {
        print!("0");
    } else {
        let ps = primes(&n);
        print!("{}", f(&n, &ps));
    }

    Ok(())
}

fn f(x: &i64, ps: &Vec<i64>) -> i32 {
    let mut y = *x;
    let mut ans: i32 = 0;
    let mut z;

    for p in ps {
        z = *p;
        while y % z == 0 {
            y /= z;
            z *= p;
            ans += 1;
        }
    }

    ans
}