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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program main
  implicit none
  integer(8) i, N, lim, m(39), t, ans, exp, level
  !integer(8),allocatable:: A(:,:)
  read(*, *) N
  lim = floor(N ** 0.5) + 1
  !allocate(A(N,2)
  m = 0
  t = 0
  do i = 1,39
    m(i) = t+i
    t = m(i)
  end do
  !read(*, *) (L(i), i = 1,N)
  t = 2
  exp = 0
  level = 0
  ans = 0
  if (N == 1) then
    ans = 0
  else
    do
      if (mod(N,t) == 0) then
        exp = exp + 1
        N = N/t
        if (exp == m(level+1)) then
          level = level + 1
        end if
      else
        ans = ans + level
        exp = 0
        level = 0
        t = t+1
      endif
      if (t == lim) then
        if (ans == 0) then
          ans = ans + 1
        endif
        exit
      endif
    end do
  endif
  write(*, *) ans
end program main