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.split.map(&:to_i)

def is_prime(p)
  n = 2
  while n * n < p
    if p % n == 0
      return false
    end
    n += 1
  end
  return true
end

n = N[0]
cnt = 0
2.upto(10**5) { |p|
  if not is_prime(p)
    next
  end
  z = 1
  1.upto(10**3) { |i|
    z *= p
    if z > n
      break
    end
    if n % z == 0
      n = n / z
      cnt += 1
      #puts z
    end
  }
  if p > n
    break
  end
}

if n > 24 and cnt == 0
  puts 1
else
  puts cnt
end