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 Common Lisp (SBCL 2.0.3) (judge verdict: Wrong Answer)

(defun prime-array (n) ;素数列
  (let ((arr (make-array (1+ n) :element-type 'bit :initial-element 1)))
    (setf (aref arr 0) 0)
    (setf (aref arr 1) 0)
    (loop :for q :from 1 :upto n :do(if (= 1 (aref arr q))
                                        (loop :for a :from (* 2 q) :upto n :by q :do(setf (aref arr a) 0))))
    arr))
(let* ((n (read))
       (pr (prime-array (ceiling (sqrt n))))
       (stk nil))
  (loop :for k :across pr
        :for a :from 0
        :if (and (= k 1) (= 0 (mod n a)))
          :do (push (cons a (loop :for p :from 1
                                  :if (not (= 0 (mod n (expt a p))))
                                    :return (1- p))) stk))
  (if stk
      (princ (loop :for k :in stk
                   :sum (loop :for a :from 1
                              :for b := (* 1/2 a (1+ a))
                              :if (< (cdr k) b)
                                :return (1- a))))
      (princ 1)))