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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

long long N;
int res = 0;

void load(void);
void calc(void);
void output(void);

int main(){
	load();
	calc();
	output();
	return 0;
}

void load(void){
	cin >> N;
}

void calc(void){
	int i, j;
	int buf;
	for(i = 2;i < 1000001;i++){
		buf = 0;
		for(j = 0;j < 100000;j++){
			if(N % i == 0){
				buf++;
				N = (long long)(N / (long long)i);
			}else{
				break;
			}
		}
		for(j = 1;j < 100000;j++){
			buf = buf - j;
			if(buf >= 0){
				res++;
			}else{
				break;
			}
		}
		if(N == 1){
			break;
		}
	}
	if(N != 1){
		res++;
	}
}

void output(void){
	cout << res;
}