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)

//
//  main.cpp
//  ABC_169_D
//
//  Created by Joe Mori on 2020/05/31.
//  Copyright © 2020 Joe Mori. All rights reserved.
//

#include <iostream>
#include<vector>
#include<queue>
bool is_prime(long long N) {
    if (N == 1) return false;
    for (long long i = 2; i * i <= N; ++i) {
        if (N % i == 0) return false;
    }
    return true;
}
int main() {
    // insert code here...
    long long int a;
    std::cin>>a;
    long long int count=0;
    long long int i;
//    std::vector<int>vec(a/2);
    i=2;
//    std::queue<int>q;
    if(is_prime(a)){
        std::cout<<1<<std::endl;
    }else{
        while(true){
        if(a%i==0){
            count++;
            a=a/i;
        }
        i++;
        if(a==1||i>a)break;
    }
    std::cout<<count<<std::endl;
    }
    return 0;
}