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: Runtime Error)

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.Scanner;
public class Main {
	public static void main(String[] args)throws Exception{
		Scanner stdIn=new Scanner(System.in);
		long N=stdIn.nextLong();
		ArrayList<Long> P=new ArrayList<Long>();
		Deque<Long> Q=new ArrayDeque<Long>();
		int ans=0,l=0;
		for(long i=2;i<=Math.sqrt(N)+1;i+=2){
			boolean key=false;
			if(N%i==0)
				if(i==2){
					P.add((long)2);i--;
					Q.add(N/2);
				}
				else{
					for(int j=0;j<P.size();j++){
						if(Math.sqrt(i)<P.get(j))
							break;
						if(i%P.get(j)==0)
							key=true;
					}
					if(key==false){
						P.add((long)i);
						Q.add(N/i);
					}
				}
			if(i==2)
				i--;
		}l=Q.size();
		for(int i=0;i<l;i++){
			boolean key=false;
			long I=Q.poll();
			for(int j=0;j<P.size();j++){
				if(Math.sqrt(I)<P.get(j))
					break;
				if(I%P.get(j)==0)
					key=true;
			}
			if(key==false){
				P.add(I);
			}
		}
		if(P.size()==0)
			P.add(N);
		for(int i=0;N>1;i++){
			long a=0,b=P.get(i);
			while(N%b==0){
				N/=b;a++;
			}
			for(int j=1;a>=j;j++){
				a-=j;ans++;
			}
		}
		System.out.println(ans);
	}
}