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::{fastout, input};
// use proconio::marker::*;

#[fastout]
fn main() {
   input! {
      mut n: usize,
   }
   let mut i = 2;
   let mut ans = 0;
   let s: Vec<_> = (1..20)
      .scan(0, |state, i| {
         *state += i;
         Some(*state)
      })
      .collect();
   while i * i < n {
      if n % i == 0 {
         let mut c = 0;
         while n % i == 0 {
            c += 1;
            n /= i;
         }
         ans += s.upper_bound(&c);
      }
      i += 1;
   }
   if n != 1 {
      ans += 1;
   }
   println!("{}", ans);
}


/// Equivalent to std::lowerbound and std::upperbound in c++
pub trait BinarySearch<T> {
   fn lower_bound(&self, x: &T) -> usize;
   fn upper_bound(&self, x: &T) -> usize;
}
#[allow(dead_code)]
impl<T: Ord> BinarySearch<T> for [T] {
   fn lower_bound(&self, x: &T) -> usize {
      let mut low = 0;
      let mut high = self.len();
      while low != high {
         let mid = (low + high) / 2;
         match self[mid].cmp(x) {
            std::cmp::Ordering::Less => {
               low = mid + 1;
            }
            std::cmp::Ordering::Equal | std::cmp::Ordering::Greater => {
               high = mid;
            }
         }
      }
      low
   }
   fn upper_bound(&self, x: &T) -> usize {
      let mut low = 0;
      let mut high = self.len();
      while low != high {
         let mid = (low + high) / 2;
         match self[mid].cmp(x) {
            std::cmp::Ordering::Less | std::cmp::Ordering::Equal => {
               low = mid + 1;
            }
            std::cmp::Ordering::Greater => {
               high = mid;
            }
         }
      }
      low
   }
}