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: Accepted)

import java.util.*;
import java.io.*;
class Main{
  public static void main(String args[]){
    long a = readLong();
    ArrayList<Long> beki = new ArrayList<Long>();
    boolean flg = true;
    for(long i=2;a>1;){
      if(a%i==0){
        a/=i;
        if(flg){
          beki.add(1L);
          flg = false;
        }else{
          beki.set(beki.size()-1,beki.get(beki.size()-1)+1);
        }
      }else{
        flg = true;
        if(i*i>a){
          beki.add(1L);
          break;
        }
        i++;
      }
    }
    long count=0;
    for(int i=0;i<beki.size();i++){
      long e = 1;
      long z = beki.get(i);
      while(z>=e){
        count++;
        z-=e;
        e++;
      }
    }
    pl(count+"");
  }
  static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  public static void pr(String str){
    System.out.print(str);
  }
  public static void pl(String str){
    System.out.println(str);
  }
  public static String readL(){
    try{
      return br.readLine();
    }catch(IOException e){
      e.printStackTrace();
      return "";
    }
  }
  public static String readS(){
    StringBuilder sb = new StringBuilder();
    while(true){
      try{
        int k = br.read();
        if(k==-1||(char)k==' '||(char)k=='\n'){break;}
        sb.append((char)k);
      }catch(IOException e){
        e.printStackTrace();
      }
    }
    return sb.toString();
  }
  public static int readI(){
    return stoi(readS());
  }
  public static long readLong(){
    return Long.parseLong(readS());
  }
  public static String[] readSs(){
    return readL().split(" ");
  }
  public static int[] readIs(){
    return stoi(readSs());
  }
  public static int stoi(String s){
    return Integer.parseInt(s);
  }
  public static int[] stoi(String s[]){
    int a[]=new int[s.length];
    for(int i=0;i<s.length;i++){
      a[i]=stoi(s[i]);
    }
    return a;
  }
  public static String itos(int i){
    return String.valueOf(i);
  }
  public static String[] itos(int[] a){
    String s[]=new String[a.length];
    for(int i=0;i<a.length;i++){
      s[i]=itos(a[i]);
    }
    return s;
  }
  public static String ctos(char c){
    return String.valueOf(c);
  }
  public static String cstos(char[] c){
    return new String(c);
  }
  public static char stoc(String s){
    return s.charAt(0);
  }
  public static char[] stocs(String s){
    return s.toCharArray();
  }
}