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(dead_code)]
#![allow(unused_imports)]
use ascii::{AsciiChar, AsciiStr, AsciiString};
use indexmap::IndexMap;
use itertools::*;
use libm::*;
use num::*;
use ordered_float::OrderedFloat;
use permutohedron::LexicalPermutation;
use proconio::marker::{Bytes, Chars};
use proconio::{fastout, input};
use std::cmp::{max, min, Ordering};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::mem::swap;

fn is_prime(n: i64) -> bool {
    if n <= 1 {
        return false;
    }

    for i in 2..=(sqrt(n as f64).ceil() as i64) {
        if n % i == 0 {
            return n / i == 1;
        }
    }
    return true;
}

fn main() {
    input!(n: i64);
    if n == 1 {
        println!("0");
        return;
    }
    let r = sqrt(n as f64).ceil() as i64;
    let mut ans = 0;
    let mut t = n;
    for z in 2..=r {
        if !is_prime(z) {
            continue;
        }
        let mut cnt = 0;
        while t % z == 0 {
            t /= z;
            cnt += 1;
        }
        let mut e = 1;
        while e <= cnt {
            cnt -= e;
            ans += 1;
            e += 1;
        }
    }
    if t == n {
        ans = 1;
    }
    println!("{}", ans);
}