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 Scala (2.13.1) (judge verdict: Wrong Answer)

object Main extends App {
  val in = new java.util.Scanner(System.in)
  val N = in.nextLong

  val divs = 0 +: (1 to 10).flatMap(i => Array.fill(i + 1)(i))
  val primes: LazyList[Int] = 2 #:: (LazyList.from(3, 2) filter {n =>
    primes takeWhile {_ <= Math.sqrt(n).toInt} forall {n % _ != 0}
  })

  def div(n: Long, ps: LazyList[Int], c: Int): Int = {
    if (n == 1L) c
    else if (ps.head >= 100000) c + 1
    else {
      def divCount(n0: Long, p: Long, c0: Int): (Long, Int) = {
        if (n0 % p == 0) divCount(n0 / p, p, c0 + 1)
        else (n0, c0)
      }
      val (n0, c0) = divCount(n, ps.head, 0)
      div(n0, ps.tail, c + divs(c0))
    }
  }

  println(div(N, primes, 0))
}