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)

def fac(num)
  return [0] if num < 2
  ans = Array.new(0)
  tmp = num
  for div in 2..(10**6)
    while tmp % div == 0
      ans.push(div)
      tmp /= div
    end
    break if tmp == 1
  end
  ans.push(tmp) if tmp > 1
  return ans
end

n = gets.to_i

a = fac(n)
fs = Hash.new(0)
a.each{|e| fs[e] += 1}

ans = 0
fs.each do |k,v|
  val = v
  cnt = 1
  while val-cnt >= cnt+1
    val -= cnt
    cnt += 1
    ans += 1
  end
  ans += 1 if val > 0
end
if n > 1
  puts ans
else
  puts 0
end