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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;

ll prime_factorization(ll a)
{
    ll i, sq;
    ll ans = 0, cnt = 0, tmp = 0;

    while (a % 2 == 0) // 偶数の場合
    {
        cnt++;
        if (cnt > tmp)
        {
            ans++;
            tmp++;
            cnt = 0;
        }
        //printf("%d\n", 2);
        a /= 2;
    }

    int flag = 0;
    sq = sqrt(a);
    for (i = 3; i <= sq; i += 2) // 3以上√a以下の奇数の場合
    {
        cnt = 0;
        tmp = 0;
        while (a % i == 0)
        {
            cnt++;
            if (cnt > tmp)
            {
                ans++;
                tmp++;
                cnt = 0;
                flag = 1;
            }
            //printf("%lld\n", i);
            a /= i;
        }
        sq = sqrt(a);
    }

    if (a != 1) // 偶数でも3以上√a以下の奇数の場合でも割り切れない場合、aが1でないなら、a自身は素数
    {
        if (flag)
        {
            cnt++;
            if (cnt > tmp)
            {
                ans++;
                tmp++;
                cnt = 0;
            }
        }
        else
            ans++;
        //printf("%lld\n", a);
    }
    return ans;
}

int main(void)
{
    ll n;
    int ans = 0;
    scanf("%lld", &n);

    printf("%lld", prime_factorization(n));
    return 0;
}