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)

/*
 * Note: This template uses some c++11 functions , so you have to compile it with c++11 flag.
 *       Example:-   $ g++ -std=c++11 c++Template.cpp
 *
 * Author : Denzil Bernard  / Edgar Monis
 * Handle: 1_mn_RMY / the_king_in_the_north
 *
 */

// C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
//#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif

using namespace std;

/*******  All Required define Pre-Processors and typedef Constants *******/
#define FOR(i, j, k, in) for (int i=j ; i<k ; i+=in)
#define RFOR(i, j, k, in) for (int i=j ; i>=k ; i-=in)
#define REP(i, j) FOR(i, 0, j, 1)
#define RREP(i, j) RFOR(i, j, 0, 1)
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.end(), cont.begin()
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define PB push_back
#define MOD 1000000007
#define read(type) readInt<type>()
#define ll long long
typedef pair< ll, ll> PII;
typedef vector<ll> VI;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<ll,ll> MPII;
typedef set<ll> SETI;
typedef multiset<ll> MSETI;

ll SI(){
    ll x;
    cin>>x;
    return x;
}


ll mod(ll a){
    return a<0?-a:a;
}

VI primeFactors(ll n)  
{  
    VI fact;
    // Print the number of 2s that divide n  
    while (n % 2 == 0)  
    {  
        // cout << 2 << " ";  
        fact.push_back(2);
        n = n/2;  
    }  
  
    // n must be odd at this point. So we can skip  
    // one element (Note i = i +2)  
    for (int i = 3; i <= sqrt(n); i = i + 2)  
    {  
        // While i divides n, print i and divide n  
        while (n % i == 0)  
        {  
            // cout << i << " ";  
            fact.push_back(i);
            n = n/i;  
        }  
    }  
  
    // This condition is to handle the case when n  
    // is a prime number greater than 2  
    if (n > 2)  
        fact.push_back(n);
        // cout << n << " ";  

    return fact;
}


/********** Main()  function **********/
int main()
{
    ll t = SI();
    VI fact = primeFactors(t);
    MPII mp;
    for(auto x: fact){
        mp[x]++;
    }
    ll ans = 0;
    for(auto x:mp){
        // cout<<x.first<<"\n";
        // cout<<x.second<<"\n";

        ll b = x.second;
        ll a = 1;
        ll num = 0;
        while(b>=a){
            b-=a++;
            num++;
        }
        ans+=num++;        
    }
    cout<<ans;

    
    return 0;
}