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)]

//input
use proconio::input;
use proconio::marker::{Bytes, Chars};
//use std::io::Read;

//collections
//use std::collections::BTreeMap;
//use std::collections::BTreeSet;
use std::collections::HashMap;
//use std::collections::HashSet;
//use std::collections::VecDeque;
//use std::collections::BinaryHeap;

//use std::cmp::*;

fn main() {
    input! {
        mut n: i64,
    }
    let mut v = vec![];

    for i in 2..=n {
        if i * i > n {
            break;
        }
        if n % i == 0 {
            v.push(i);
        }
        if n / i != 0 {
            v.push(n / i);
        }
    }
    v.sort();

    //println!("{:?}", v);
    let mut cnt = 0u64;
    let mut i = 0usize;
    //println!("{}", v[v.len() - 1]);
    while n != 1 {
        if i == v.len() {
            break;
        }
        if n % v[i] == 0 {
            cnt += 1;
            n /= v[i];
        }
        i += 1;
    }
    if n != 1 && cnt == 0 {
        cnt = 1;
    }
    println!("{}", cnt);
}