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: Wrong Answer)

// 問題の URL を書いておく
// 

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <bitset>
#include <numeric>
using namespace std;

//#define ENABLE_PRINT

template<typename T>
T Input()
{
    T v;
    cin >> v;
    return v;
}

template<typename T>
vector<T> InputVector(int n)
{
    vector<T> v(n);
    for(int i = 0; i < n; ++i)
    {
        cin >> v[i];
    }
    return v;
}

template<typename T>
void InputArray(T* begin, int n)
{
    for(int i = 0; i < n; ++i)
    {
        cin >> begin[i];
    }
}

#if defined(ENABLE_PRINT)

#define Print(v) \
do {\
    cout << #v << ": " << v << endl; \
}while(0)

#define PrintVec(v) \
do {\
    for(int __i = 0; __i < v.size(); ++__i) \
    { \
        cout << #v << "[" << __i << "]: " << v[__i] << endl; \
    }\
}while(0)

#else

#define Print(v) ((void)0)
#define PrintVec(v) ((void)0)

#endif

int main(int, const char**)
{
    uint64_t n;
    cin >> n;
    if(n == 1)
    {
        cout << 0 << endl;
        return 0;
    }
    vector<uint64_t> divs;

    for(int i = 1; i * i <= n; ++i)
    {
        if(n % i == 0)
        {
            divs.push_back(i);
#if 0
            if(n / i != i)
            {
                divs.push_back(n / i);
            }
#endif
        }
    }

    if(divs.size() == 1)
    {
        cout << 1 << endl;
    }
    else
    {
        auto nn = n;
        int count = 0;
        auto it = divs.begin() + 1;
        while(it != divs.end())
        {
//            cout << *it << endl;
            if((nn % *it) == 0)
            {
                nn /= *it;
                count++;
            }
            it++;
        }
        cout << count << endl;
    }

    return 0;
}