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(){
    int i, j, k;
    long N;
    scanf("%ld", &N);
    long p[100000];
    i = 0;
    p[0] = 2;
    for (j = 3; j <= N;j++){
        for (k = 0; k <= i;k++){
            if(j%p[k]==0){
                break;
            }else if(k==i){
                p[++i] = j;
            }
        }
    }
    p[i+2] = 0;
    i = 0;
    int cnt_p = 0;
    while(p[i]){
        cnt_p++;
        i++;
    }
    int fac[cnt_p];
    for (i = 0; i < cnt_p;i++){
        fac[i] = 0;
    }
        i = 0;
    while(p[i]){
        if(N%p[i]==0){
            N = N / p[i];
            fac[i]++;
        }else{
            i++;
        }
    }
    int cnt = 0;
    for (i = 0; i < cnt_p;i++){
        for (j = 0; fac[i] - j > 0;j++){
            cnt++;
        }
    }
    printf("%d", cnt-1);
    return 0;
}