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)

use proconio::{fastout, input};
use std::collections::HashMap;

#[fastout]
fn main() {
    input! {in_n: u64};
    let mut n = in_n;
    let mut nums = HashMap::new();

    while n > 1 {
        for i in 2..n + 1 {
            if i * i > n {
                if nums.contains_key(&n) {
                    nums.insert(n, nums[&n] + 1);
                } else {
                    nums.insert(n, 1);
                }

                n = 1;
                break;
            }

            if n % i != 0 {
                continue;
            }

            if nums.contains_key(&i) {
                nums.insert(i, nums[&i] + 1);
            } else {
                nums.insert(i, 1);
            }

            n = n / i;

            break;
        }
    }

    let mut c = 0;

    for (_, nv) in &nums {
        let mut v = *nv;
        let mut p = 1;

        while v >= p {
            c += 1;
            v -= p;
            p += 1;
        }
    }

    println!("{}", c);
}