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 Ruby (2.7.1) (judge verdict: Wrong Answer)

require 'prime'
class Integer
    def my_divisor_list2
        return [1] if self == 1
        Prime.prime_division(self).map do |e|
        Array.new(e[1]+1).map.with_index do |element, i|
            e[0]**i
        end
        end.inject{|p,q| p.product(q)}.map do |a|
        [a].flatten.inject(&:*)
        end.sort
    end
end
  
n = gets.to_i
divs = n.my_divisor_list2
# divs = Prime.each(n).to_a 
count = 0
divs.each do |d|
    if (d.my_divisor_list2.length > 3 || d == 1) && d != 8
        next
    end
    n /= d
    count += 1 
    break if n == 1
end
puts count