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)

N = gets.to_i

require "prime"

LIMIT = 10 ** 6 * 2

zmap1 = {}
zmap2 = {}
Prime.each(LIMIT) do |p|
  zmap1[p] = true
  e = 2
  while (z = p ** e) < LIMIT
    zmap2[z] = true
    e += 1
  end
end

zs1 = zmap1.keys.sort.reverse
zs2 = zmap2.keys.sort

n = N
count = 0

(zs1 + zs2).each do |z|
  if n > LIMIT && Prime.prime?(n)
    count += 1
    break
  end

  if n % z == 0
    count += 1
    n = n.div(z)
  end
end

puts count