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'
n = gets.to_i

pd1 = n.prime_division
pd2 = []
pd1.each do |pd|
  pd2 << [pd[0]]*pd[1]
end
pd2.flatten!

if pd2.size == 0
  puts '0'
  exit
elsif pd2.size == 1
  puts '1'
  exit
end

old = []
count = 0
tmp = 0
ok = true

pd2.each.with_index do |pd, i|
  tmp = pd if ok
  if n % tmp == 0 && !old.include?(tmp)
    old << pd
    count += 1
    n /= pd
    ok = true
  else
    tmp *= pd
    ok = false
  end
end
p count