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)]
#![allow(unused_variables)]
use proconio::input;
use proconio::marker::*;
use std::cmp::{min, max};
use std::collections::{HashSet, HashMap, BinaryHeap};

fn main() {
    input! {
        n: u64,
    }
    let mut prime = vec![];
    let mut is_prime = vec![true; 1e6 as usize + 1];
    let mut t = 2u64;
    while t * t <= n{
        if is_prime[t as usize]{
            prime.push(t);
        }
        for i in ((t * 2)..=(1e6 as u64)).step_by(t as usize){
            is_prime[i as usize] = false;
        }
        t += 1;
    }
    let mut ans = 0;
    let mut t = n;
    for &i in prime.iter(){
        let mut cnt = 0;
        while t % i == 0{
            cnt += 1;
            t /= i;
        }
        let mut u = 1;
        while cnt - u >= 0{
            ans += 1;
            cnt -= u;
            u += 1;
        }
    }
    if n > 1{
        ans += 1;
    }
    println!("{}", ans);
}