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)

// -*- coding:utf-8-unix -*-

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

fn main() {
    input! {
        mut n: u64,
    }

    let mut hm = HashMap::new();
    let mut divisor = 2;
    while n > 1 {
        // println!("n, divisor: {}, {}", n, divisor);
        if n % divisor == 0 {
            let counter = hm.entry(divisor).or_insert(0);
            *counter += 1;
            n /= divisor;
        } else if divisor * divisor > n {
            let counter = hm.entry(n).or_insert(0);
            *counter += 1;
            break;
        } else {
            divisor += 1;
        }
    }
    // println!("{:?}", hm);

    let mut ans = 0;
    for (_, v) in hm.iter() {
        // println!("v: {}", v);
        let mut a = *v;
        let mut b = 1;
        while a >= b {
            a -= b;
            b += 1;
            ans += 1;
        }
    }
    println!("{}", ans);
}