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>
#include <string>
#include <math.h> 
#include <iomanip>
#include <vector>
#include <map>
using namespace std;

std::map<unsigned long long,unsigned long long> m;
int answer = 0;
void cal(unsigned long long n){
    int a = 2;
    
    while (n >= a * a) {
        if (n % a == 0) {
            if(m.count(a) >= 1){
                m[a] += 1; 
            }
            else{
                m[a] = 1;
            }
            n /= a;
        } 
        else {
            if(a==2){
                a++;
            }
            else{
                a+=2;
            }
        }
    }
    if(m.count(n) >= 1){
        m[n] += 1;
    }
    else{
        m[n] = 1;
    } 
}

int main(){
    unsigned long long N;
    cin >> N;
    int ans = 0;
    //soinsu bunkai
    if (N == 1){
        cout << 0 << endl;
        return 0;
    }
    cal(N);
    //cout << m[2] << endl;

    for(auto i = m.begin(); i != m.end() ; ++i ){
        unsigned long long dup = 1;
        unsigned long long omomi = 1;
        //cout << i->second << endl;
        while(i->second >= dup){
            ans+=1;
            ++omomi;
            dup = dup+omomi;
        }
    }

    cout << ans << endl;
    return 0;
}