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: Accepted)

import java.util.*;
import java.io.PrintWriter;
import static java.lang.Integer.*;
import static java.lang.Long.*;
import static java.lang.Math.*;
import static java.lang.System.*;

public class Main {
	public static final int MOD = 1000000007;
	public static void main(String[] args) {
		int i,j;
		Scanner sc = new Scanner(in);
		long n = parseLong(sc.next());
		sc.close();
		ArrayList<Long> pri = new ArrayList<>();
		pri.add(2L);
		for(long k=3; k*k<=n; k+=2) {
			boolean f=true;
			for (Long l : pri) {
				if(l*l>k) {
					break;
				}
				if(k%l==0) {
					f=false;
					break;
				}
			}
			if(f) {
				pri.add(k);
			}
		}
//		out.println(pri.toString());
		long ans=0;
		for (Long l : pri) {
			long cnt=0;
			while(n%l==0) {
				n/=l;
				cnt++;
			}
			if(cnt>0) {
//				out.printf("%d:%d:%d%n", l, cnt, (long)((1+Math.sqrt(1+4*cnt))/2));
				ans+=(long)((-1+Math.sqrt(1+8*cnt))/2);
			}
			if(n==1) break;
		}
//		out.println(n);
		if(n>1) ans++;
		out.println(ans);
	}
}