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)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;
use im_rc::HashMap;

#[fastout]
fn main() {
    input! {
    n: usize
    }
    if n == 1 {
        println!("0");
        return;
    }
    let mut l = vec![];
    let mut t = 2;
    let mut k = n.clone();
    while t * t <= n {
        while k % t == 0 {
            l.push(t);
            k /= t;
        }
        t += 1;
    }
    if l.is_empty() {
        l.push(n);
    }
    let mut d = 0;
    let mut ans = 0;
    let m = l.len();
    if m == 1{
        println!("1");
        return;
    }
    for i in 0..m - 1 {
        if i + d < m && l[i] == l[i + d] {
            ans += 1;
            d += 1;
        }
        if l[i] != l[i + 1] {
            d = 0;
        }
    }
    if l[m - 1] != l[m - 2] {
        ans += 1;
    }
    println!("{}", ans);
}