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::input;

mod Prime {
    pub fn is_prime(n: i64) -> bool {
        if n == 1 {
            return false;
        }
        for i in 2..n {
            if i * i > n {
                break;
            }
            if n % i == 0 {
                return false;
            }
        }
        return true;
    }

    pub fn prime_division(n: i64) -> Vec<(i64, i64)> {
        let mut v = n;
        let mut pf = Vec::new();
        if v == 0 {
            return pf;
        }
        if v < 0 {
            v = -v;
            pf.push((-1, 1));
        }
        for p in 2..n {
            if p * p > n {
                break;
            }
            let mut count = 0;
            while v % p == 0 {
                v /= p;
                count += 1;
            }
            if count > 0 {
                pf.push((p, count));
            }
            if v <= p {
                break;
            }
        }

        if v > 1 {
            pf.push((v, 1));
        }
        return pf;
    }
}

fn main() {
    input! {
        n: i64,
    }

    let pf = Prime::prime_division(n);
    let ans = pf.into_iter().fold(0, |acc, (base, exp)| {
        if base == 1 {
            acc
        } else {
            let mut x = exp;
            let mut count = 1;
            while x >= count {
                x -= count;
                count += 1;
            }
            acc + count - 1
        }
    });

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