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 Haskell (GHC 8.8.3) (judge verdict: Time Limit Exceeded)

import qualified Data.ByteString.Char8 as BS
import Data.List
import Data.Maybe

readInt = fst . fromJust . BS.readInteger

getInt = readInt <$> BS.getLine

factors :: Integer -> [Integer]
factors n = [x | x <- [1 .. n], n `mod` x == 0]

factorization :: Integer -> [Integer]
factorization 1 = []
factorization x = v : factorization (x `div` v)
  where
    v = factors x !! 1

countNum :: [Integer] -> [Int]
countNum [] = []
countNum xs = k : countNum xs2
  where
    num = head xs
    (xs1, xs2) = span (== num) xs
    k = length xs1

checkParts :: Int -> Int -> Int
checkParts t n
  | t >= n = 1 + checkParts (t - n) (n + 1)
  | otherwise = 0

calc :: [Int] -> Int
calc [] = 0
calc (t : ts) = checkParts t 1 + calc ts

main = do
  n <- getInt
  if n == 1
    then print 0
    else print $ calc (countNum (factorization n))