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 <iostream>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
 
using namespace std;
typedef long long int ll;
typedef pair <int,int> pii;
typedef pair<ll,ll> pll;
 
 
/*bool compare_function(const pair<int, int> a, const pair<int, int> b)
{
    return a.first < b.first;
}*/
// use case
//sort(V.begin(), V.end(), compare_function);
 
/*struct compare
{
    bool operator ()(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.second-a.first==b.second-b.first)
    {
        return a.first>b.first;
    }
    return a.second-a.first < b.second-b.first;}
};
priority_queue <pair<int,int>, vector <pair<int,int>>, compare> Q;*/
 
/*void print1(vector <int> X)
{
    for (int i=0; i<X.size(); i++)
    {
        cout<<X[i]<<" ";
    }
    cout<<endl;
    return;
}
 
void print2(vector <vector <int>> X)
{
    for (int i=0; i<X.size(); i++)
    {
    for (int j=0;j<X[i].size();j++) {
        cout<<X[i][j]<<" ";}
    cout<<endl;}
    return;
}*/


 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin>>n;
    ll p=2;
    int ats=0;
    while (p*p<=n) {
		if (n%p==0) {
			int laip=0;
			while (n%p==0) {n=n/p; laip++;}
			int i=1;
			while (laip>=i) {ats++; laip=laip-i; i++;}
			}
			p++;
		}
		if (n>1) {ats++;}
		cout<<ats<<endl;
    return 0;
}