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)

use proconio::*;

fn main() {
    input!{
        mut n: usize,
    }

    let mut is_prime = vec![true; 1000010];
    is_prime[0] = false;
    is_prime[1] = false;


    let mut ans = 0;
    for i in 2..1000010 {
        if is_prime[i] {
            for j in (i*i..1000010).skip(i) {
                is_prime[j] = false;
                let mut tmp = 0;
                while n % i == 0 {
                    tmp += 1;
                    n /= i;
                }

                for k in 1..50 {
                    if tmp >= k {
                        ans += 1;
                        tmp -= k;
                    } else {
                        break;
                    }
                }
            }
        }
    }

    if n > 1 {
        ans += 1;
    }

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