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: Accepted)

# frozen_string_literal: true

def factor(n)
  return [1] if n == 1

  `factor #{n}|sed -r 's/[0-9]+://g'|tr ' ' '\n'|sort -n`.split.map(&:to_i)
end

# Example: factors(100) => {2=>2, 5=>2}
def factors_and_num(n)
  factors = factor(n)
  ret = Hash.new(0)
  factors.each do |i|
    ret[i] += 1
  end
  ret
end

def uniq_factor(n)
  `factor #{n}|sed -r 's/[0-9]+://g'|tr ' ' '\n'|sort -n|uniq`.split.map(&:to_i)
end

n = gets.to_i

if n == 1
  puts 0
  exit
end

f = factors_and_num(n)

ans = 0

f.each do |_k, v|
  s = 0
  count = 1
  while v - count >= 0
    s += 1
    v -= count
    count += 1
  end
  ans += s
end

puts ans