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 Julia (1.4.0) (judge verdict: Wrong Answer)

MAXP = 10^6
f = [i for i in 1:MAXP]
p = []
for i in 2:MAXP
  MAXP < i^2 && break
  if f[i] == i
    push!(p, i)
    j = 2i
    while j <= MAXP
      f[j] == j && (f[j] = i)
      j += i
    end
  end
end

function factor(n)
  fs = Dict{Int64, Int}()
  for pi in p
    n <= MAXP && break
    while n % pi == 0
      !haskey(fs, pi) && (fs[pi] = 0)
      fs[pi] += 1
      n ÷= pi
    end
  end
  MAXP < n && (fs[n] = 1; n = 1)
  while 1 < n
    !haskey(fs, f[n]) && (fs[f[n]] = 0)
    fs[f[n]] += 1
    n ÷= f[n]
  end
  return fs
end

function s(n)
  s = 0
  for v in values(factor(n))
    i = 1
    k = 0
    while k + i <= v
      s += 1
      k += i
      i += 1
    end
  end
  return s
end

n = parse(Int64, readline())
println(s(n))