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)]

use proconio::input;
use proconio::marker::{Bytes, Chars, Usize1};
use std::cmp::{Reverse, min, max};
use itertools::Itertools;

fn isqrt(n: i64) -> i64 {
    (n as f64).sqrt() as i64
}

#[proconio::fastout]
fn main() {
    input! {
        n: i64,
    }

    if n == 1 {
        println!("0");
        return;
    }

    let mut m = n;
    let mut ans = 0;
    for i in 2.. {
        if i * i > n {
            break
        }

        let mut e = 0i64;
        while m % i == 0 {
            m /= i;
            e += 1;
        }
        for k in 1.. {
            if e >= k {
                ans += 1;
                e -= k
            } else {
                break
            }
        }
    }

    if ans == 0 {
        println!("1");
    } else {
        println!("{}", ans);
    }
}