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 <stdlib.h>
#include <math.h>

int *prime(const int max)
{
    int c, cc;
    int *ans;
    ans = (int *)malloc(sizeof(int) * (max + 1));
    for (c = 0; c <= max; c += 2)
    {
        ans[c] = 0;
    }
    for (c = 3; c <= max; c += 2)
    {
        ans[c] = c;
    }
    ans[2] = 2;
    for (c = 3; c <= sqrt(max) + 1; c += 2)
    {
        if (ans[c] == 0)
        {
            continue;
        }
        for (cc = 2; cc * c <= max; ++cc)
        {
            ans[c * cc] = 0;
        }
    }
    return ans;
}

int main(void)
{
    int *key;
    int c, ans = 0, power;
    long N;
    scanf("%ld", &N);
    key = prime((int)sqrt(N) + 3);
    for (c = 2; c <= (int)sqrt(N) + 2; ++c)
    {
        if (key[c] != 0)
        {
            power = 1;
            while (N % (long)(pow(c, power) + 0.1) == 0)
            {
                ++ans;
                N /= (long)(pow(c, power) + 0.1);
                ++power;
            }
        }
    }
    if (N != 1 && ans == 0)
    {
        ++ans;
    }
    printf("%d", ans);
    return 0;
}