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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define endl '\n'
#define _USE_MATH_DEFINES
using namespace std;

int main() {
	
	ios::sync_with_stdio(false);
    cin.tie(0);
    
    long long int store[20][2];
    long long int n;
    int times = 0,pos=0;
    cin>>n;
    for(int x=2;x<=n;x++)
    {
    	if(n%x==0){
    		int much = 0;
    		while(n%x==0)
        	{
        		much++;
            	store[pos][0] = x;
            	n/=x;
        	}
        	store[pos][1] = much;
        	pos++;
		}
    }
  	for(int i=0;i<pos;i++){
  		int t = 1;
  		while(1){
  			if(store[i][1]>=t){
  				store[i][1]-=t;
  				times++;
  				t++;
			}
			else break;
		}
  		
	}
    cout<<times;
    
	return 0;
}