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.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		long n=scn.nextLong();
//		if(n==1){
//			System.out.println(0);
//			return;
//		}
		long val = n;
		long count=0;
		long valx= (long)Math.sqrt(n);
		for(long i=2;i*i<=val;i++){
			//int j=1;
			long val2=i;
			while(n%val2==0 && n>0){
		//	System.out.println(val2+"A"+n);
				n/=val2;
				val2*=i;
				count++;
				
			}
			while(n%i==0){
				n/=i;
			}
			if(n==0){
				break;
			}
		}
		//boolean ch=isprime(n);
		//System.out.println(n);
		count+=n>valx?1:0;
		System.out.println(count);

	}
//	public static boolean isprime(long n){
//		for(int i=2;i*i<=n;i++){
//			if(n%i==0){
//				return false;
//			}
//		}
//	}

}