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, input};
use std::collections::BTreeSet;

#[fastout]
fn main() {
    input! {
        mut n: u64
    }

    let mut primes = BTreeSet::new();

    for i in 2..=n.isqrt() + 1 {
        if n % i == 0 {
            primes.insert(i);
        }
    }
    if primes.is_empty() {
        primes.insert(n);
    }

    let ans = if n == 1 {
        0
    } else {
        let mut count = 0;
        for p in primes {
            let mut tmp = p;
            while n % tmp == 0 {
                count += 1;
                n /= tmp;
                tmp *= p;
            }
        }
        count
    };

    println!("{}", ans);
}

trait IntegerSqrt {
    fn isqrt(&self) -> Self;
}

impl IntegerSqrt for u64 {
    fn isqrt(&self) -> Self {
        if *self < 2 {
            *self
        } else {
            let s = (*self >> 2).isqrt() << 1;
            let l = s + 1;
            if l * l > *self {
                s
            } else {
                l
            }
        }
    }
}