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)

#[allow(unused)]
use proconio::marker::*;
use proconio::{fastout, input};
#[allow(unused)]
use std::cmp::{max, min};
#[allow(unused)]
use std::collections::{HashMap, HashSet, VecDeque};

#[fastout]
fn main() {
    input! {
        n: i64,
    }
    let mut rec = vec![];
    factorization(n, &mut rec);

    let mut ans = 0;
    for r in rec {
        ans += help(r);
    }
    println!("{}", ans);
}

fn factorization(n: i64, rec: &mut Vec<i64>) {
    fn factor_sub(n: i64, m: i64) -> (i64, i64) {
        let mut c = 0;
        let mut x = n;
        while x % m == 0 {
            c += 1;
            x /= m;
        }
        (c, x)
    }

    let (c, n) = factor_sub(n, 2);
    if c > 0 {
        rec.push(c);
    }
    let mut x = 3;
    let mut m = n;
    while x * x <= m {
        let (c, n) = factor_sub(m, x);
        if c > 0 {
            rec.push(c);
        }
        m = n;
        x += 2;
    }
    if m > 1 {
        rec.push(1);
    }
}

fn help(n: i64) -> i64 {
    if n <= 2 {
        1
    } else if n <= 5 {
        2
    } else if n <= 9 {
        3
    } else if n <= 14 {
        4
    } else if n <= 20 {
        5
    } else if n <= 27 {
        6
    } else if n <= 35 {
        7
    } else if n <= 44 {
        8
    } else {
        9
    }
}