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 1.8.0) (judge verdict: Time Limit Exceeded)


import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        long num = Long.parseLong(br.readLine());
        System.out.println(solve(num));
    }

    private static long solve(long num){
        if(num==1)
            return 0;
        if(isPrime(num))
            return 1;
        long prime = 2;
        int count=0;
        while(prime<=num){
            long curr = prime;
            while(num%curr==0){
                num/=curr;
                curr*=prime;
                count++;
            }
            while(num%prime==0){
                num/=prime;
            }
            prime++;
        }
        return count;
    }

    private static boolean isPrime(long num){
        for(long n=2;n*n<=num;n++){
            if(num%n==0)
                return false;
        }
        return true;
    }

}