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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const int inf = 1000000000;



int main(){
    ll n;
    cin >> n;
    if(n==1){
        cout << 0 << endl;
        return 0;
    }
    int cnt = 0;
    vector<int>prime(150000000);
    rep(i,prime.size()){
        prime[i]=1;
    }
    for(int i=2; i<sqrt(n); i++){
        if(n%i == 0){
            for(int j=2; i*j<sqrt(n); j++){
                prime[i*j]=0;
            }
        }
    }
    int size = prime.size();
    for(int i=2; i<size; i++){
        if(prime[i] == 1){
            for(int j=1; j<10000000; j++){
                int a = pow(i,j);
                if(n%a==0){
                    //cout << a << endl;
                    cnt++;
                    n/=a;
                }
                else{
                    break;
                }
            }
        }
    }
    if(cnt == 0){
        cout << 1 << endl;
        return 0;
    }

    cout << cnt << endl;
}