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)

/*
    @uthor: Amit Kumar
    user --> GitHub: drviruses ; CodeChef: dr_virus_ ; Codeforces,AtCoder,Hackerearth,Hakerrank: dr_virus;
*/
#include <bits/stdc++.h>
#include <chrono>
using namespace std;
using namespace std::chrono;
//#include <boost/multiprecision/cpp_int.hpp>
//namespace mp = boost::multiprecision;
//#define ln mp::cpp_int;
#define int    long long
#define ld    long double
#define ull   unsigned long long
#define endl "\n"

vector<int> arr;
void factorize(long long n) 
{ 
    int count = 0; 
  
    // count the number of times 2 divides  
    while (!(n % 2)) { 
        n >>= 1; // equivalent to n=n/2; 
        count++; 
    } 
  
    // if 2 divides it 
    if (count) 
        arr.push_back(count);
  
    // check for all the possible numbers that can  
    // divide it 
    for (long long i = 3; i <= sqrt(n); i += 2) { 
        count = 0; 
        while (n % i == 0) { 
            count++; 
            n = n / i; 
        } 
        if (count) 
            arr.push_back(count);
    } 
  
    // if n at the end is a prime number. 
    if (n > 2) 
        arr.push_back(1);
} 

void virus(){
        int n;
        cin>>n;
        factorize(n);
        int pre[100000];
        pre[0]=1;
        for(auto i=1;i<100000;i++){
            pre[i] = pre[i-1] + (i+1);
        }
        int cnt = 0;
        for(auto i:arr){
            //cout<<i<<endl;
            cnt += (upper_bound(pre,pre+100000,i) - pre);
        } 
        cout<<cnt;
}


int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
  /*#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif*/

    int t = 1;
    //cin>>t;
    while(t--){
           auto start = high_resolution_clock::now();
           virus();
           auto stop = high_resolution_clock::now();
           auto duration = duration_cast<seconds>(stop - start);
         //cout << "Time: "<<duration.count()<<endl;
        //your code goes here
    }
    return 0;
}