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: i64,
    }

    let mut ans = 0;
    for (_, count) in factor(n) {
        let mut n = count;
        let mut b = 1;
        while b <= n {
            n -= b;
            b += 1;
            ans += 1;
        }
    }
    println!("{}", ans);
}

// O(n^(1/2)) で与えられた値の素因数分解を求める
use std::collections::HashMap;

fn factor(num: i64) -> HashMap::<i64, i64> {
    let mut map = HashMap::new();
    let mut n = num;

    let mut i = 2;
    loop {
        if n == 1 { 
            // 割り切れたら抜ける
            break;
        }
        if i * i >= num {
            // n^(1/2)まで試す
            // 残っていたら追加して抜ける
            map.insert(n, 1);
            break;
        }
        if n % i == 0 {
            // Mapに無ければ追加、あれば1増やす
            *map.entry(i).or_insert(0) += 1;
            n /= i;
            continue;
        }
        i += 1;
    }
    map
}