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;
fn main() {
    input!(n: usize);
    let mut ans = 0;
    let table = trial_division(n);
    //println!("{:?}", table);
    for (&a, &i) in &table {
        if a == 1 {
            continue;
        }
        //let tmp = calc(i);
        //println!("{} => {}", i, tmp);
        ans += calc(i);
    }

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

fn calc(n: usize) -> usize {
    for i in 1..n + 1 {
        if (i * (i + 1)) / 2 == n {
            return i;
        } else if (i * (i - 1)) / 2 < n && n < (i * (i + 1)) / 2 {
            return i - 1;
        }
    }
    return 0;
}

// 試し割り
// O(√N)
use std::collections::HashMap;
fn trial_division(n: usize) -> HashMap<usize, usize> {
    let mut primes = HashMap::new();
    let mut i = 2;
    let mut n = n;

    //  n を root n 以下の整数で割り切れるまで割っていく
    while i * i <= n {
        while n % i == 0 {
            n /= i;
            // match primes.get_mut(&i) {
            //   None => { primes.insert(i, 1); },
            //   Some(x) => {
            //       *x += 1;
            //   },
            // }

            match primes.get(&i) {
                Some(&val) => {
                    primes.insert(i, val + 1);
                }
                _ => {
                    primes.insert(i, 1);
                }
            }
        }
        i += 1;
    }

    // 最後にnが素数になっている場合はそれ自身も素因数に含めて終了
    // match primes.get_mut(&n) {
    //   None => { primes.insert(n, 1); },
    //   Some(x) => {
    //     *x += 1;
    //   },
    // }

    match primes.get(&n) {
        Some(&val) => {
            primes.insert(n, val + 1);
        }
        _ => {
            primes.insert(n, 1);
        }
    }

    primes
}