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::input;
use std::collections::HashMap;

fn main() {
    input! {
        n: u64,
    }
    if n == 1 {
        println!("{}", 0);
        return;
    }
    println!(
        "{:?}",
        primal_factorization(n)
            .values()
            .map(|&v| divival(v))
            .fold(0, |sum, x| sum + x)
    );
}

fn primal_factorization(n: u64) -> HashMap<u64, u64> {
    let mut map = HashMap::new();
    let (mut prime, mut remain) = (1, n);
    while prime * prime <= n {
        prime += 1;
        while remain % prime == 0 {
            remain /= prime;
            map.entry(prime).and_modify(|e| *e += 1).or_insert(1);
        }
    }
    if map.is_empty() {
        map.insert(n, 1);
    }
    map
}

// 1 + 2 + ... + n = n(n+1)/2
// n(n-1)/2 < x < n(n+1)/2
fn divival(x: u64) -> u64 {
    let mut i = 1;
    while !(i * (i - 1) / 2 <= x && x < i * (i + 1) / 2) {
        i += 1
    }
    i - 1
}