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: Runtime Error)


import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {
    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        long n = sc.nextLong();

        if(n == 1){
            System.out.println("0");
            return;
        }

        //如果一开始就是质数，那就return
        if(isPrime(n)){
            System.out.println("1");
            return;
        }

        int ans = 0;
        Set<Long> zSet =  new HashSet<>();

        List<Long> primes = findPrime2(1000000L);
        //如果一开始不是质数，先做出质数列表
        for (int i = 0; i < n; i++){
            long p = primes.get(i);
            for(int j = 1; j < n; j++){
                long z = (long) Math.pow(p, j);
                if(z>n){
                    break;
                }
                if(n%z==0){
                    //System.out.println("z = " + z);
                    n = n/z;
                    zSet.add(z);
                    //System.out.println("n = " + n);
                    ans ++;
                    if (isPrime(n)){
                        if(zSet.contains(n) || n==1){
                            System.out.println(ans);
                            return;
                        }else {
                            System.out.println(ans+1);
                            return;
                        }
                    }
                }
            }
        }
    }

    public static boolean isPrime(long n){

        for (long i=2;i<=(int)Math.sqrt(n);i++){
            if (n%i==0) return false;
        }
        return true;
    }


    public static List<Long> findPrime2(long n) {
        List<Long> primes = new ArrayList<>();
        primes.add(2L);
        for(long i = 3; i <= n; i++) {
            long tmp = (long) Math.sqrt(i) + 1;
            for(long j = 2; j <= tmp; j++) {
                if(i % j == 0) {
                    break;
                }
                if(j == tmp){
                    primes.add(i);
                }
            }
        }
        return primes;
    }

}