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: Wrong Answer)

#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a,const void *b){
    if(*(long long *)a>*(long long *)b)return 1;
    return -1;
}
int main(int argc, const char * argv[]) {
    long long n,ans=0;
    scanf("%lld",&n);
    if(n==1){
        printf("0");
        return 0;
    }
    long long CN=n;
    for(int i=2;i<=sqrt(CN)+1;i++){
        if(n%i==0){
            int cnt=0;
            while(n%i==0){
                n/=i;
                cnt++;
            }
            for(int j=1;cnt>=j;j++){
                cnt-=j;
                ans++;
            }
            
        }
    }
    if(ans==0)ans++;
    printf("%lld",ans);
    
    return 0;
}