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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
public class Main{
	static ArrayList<Integer> primes = new ArrayList<Integer>();
   public static void main(String[] main) throws Exception{
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st = new StringTokenizer(br.readLine());
      PrintWriter out = new PrintWriter(System.out);
      long n = Long.parseLong(st.nextToken());
      sieve((int)Math.sqrt(n));
      int[] triangle = new int[] {1, 3, 6, 10, 15, 21, 28, 36, 45};
      int ans = 0;
      for(int p: primes) {
    	  int pow = 0;
    	  while(n%p == 0) {
    		  n /= p;
    		  pow++;
    	  }
    	  int i = 0;
    	  for(; i < 10; i++)
    		  if(triangle[i] > pow)
    			  break;
    	  ans += i;
    	  if(n == 1)
    		  break;
      }
      if(ans == 0)
    	  ans++;
      out.println(ans);
      out.close();
   }
   static void sieve(int n) {
	   boolean[] dp = new boolean[n+1];
	   for(int i = 2; i < n; i++)
		   dp[i] = true;
	   for(int p = 2; p*p <= n; p++) {
		   for(int i = p*p; i <= n; i+= p) {
			   dp[i] = false;
		   }
	   }
	   for(int i = 2; i < n; i++)
		   if(dp[i])
			   primes.add(i);
   }
}