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: Accepted)

#include<bits/stdc++.h>
#define MAX 1000001
using namespace std;
void frac_init(vector<int> &frac){
    int flag;
    int n[MAX];
    for(int i=0; i<MAX; i++){
        n[i]=1;
    }
    for(int num=2; num<MAX; num++){
        if(n[num]==1){
            frac.push_back(num);
            for(int i=num; i<MAX; i+=num){
                n[i]=0;
            }
        }
    }
}
int main(){
    long long N,Norig;
    int stack[41],s;
    int f,e,ans=0;
    vector<int> frac;
    frac_init(frac);
    stack[0]=0;
    for(int i=0; i<9; i++){
        for(int j=(i*(i+1))/2; j<((i+1)*(i+2))/2; j++){
            if(j>40) break;
            stack[j]=i;
        }
    }
    cin >> N;
    Norig=N;
    for(int i=0; i<frac.size(); i++){
        f=frac.at(i);
        e=0;
        while(N%f==0){
            N/=f;
            e++;
        }
        ans+=stack[e];
        if(N==1) break;
    }
    if(N>1) ans++;
    cout << ans << endl;
    return 0;
}