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 <stdio.h>
#include <math.h>

int main(void){

    long N = 0;
    int count = 0;
    int count2 = 0;
    int check = 0;
    scanf("%ld", &N);
    long tmpN = 0;
    long A[1000];
    tmpN = N;
    
    for (int i = 1; (double)i <= sqrt((double)N); i++){
        if (N % i == 0){
            A[count] = i;

            if (tmpN % A[count] == 0 && i != 1){
                tmpN = tmpN / A[count];
                count2++;
                check = 1;
            }
            count++;
        }
    }
    if (N == 1){
        printf("0\n");
    } else if (check == 0){
        printf("1\n");    
    } else {
        printf("%d\n", count2);
    }

    return 0;
}