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 <iostream>

using namespace std;
const int lim = 1e7;
int lp[lim + 10];
long long n;
void sieve(){
    for (long long i = 2; i * i <= lim; i++)
        if (lp[i] == 0){
            lp[i] = 1;
            for (long long j = i; j <= lim; j *= i)
                lp[j] = 1;
        }
}
bool prime(long long x){
    if (x < 2) return false;
    for (long long i = 2; i * i <= x; i++)
        if (x % i == 0) return false;
    return true;
}
int main(){
    sieve();
    cin >> n;
    long long cnt = 0 , temp = 2;
    for (int i = 2; i <= lim; i++)
        if (lp[i] == 1 && n % i == 0) n /= i , cnt++;
    if (prime(n)) cnt++;
    cout << cnt;
    return 0;
}