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)

#include<iostream>
#include <bits/stdc++.h>
using namespace std;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define ld long double
#define mp make_pair
#define pii pair<int,int>
#define mod 1000000007
#define rep(i,n) for(int i=0;i<n;i++)
#define MAXN 100
  bool isPrime(ll n) 
{ 
    // Corner cases 
    if (n <= 1)  return false; 
    if (n <= 3)  return true; 
  
    // This is checked so that we can skip  
    // middle five numbers in below loop 
    if (n%2 == 0 || n%3 == 0) return false; 
  
    for (ll i=5; i<=sqrt(n); i=i+6) 
        if (n%i == 0 || n%(i+2) == 0) 
           return false; 
  
    return true; 
} 
// stores smallest prime factor for every number 
ll primeFactors(ll n)  
{  
    ll c=0;
    // Print the number of 2s that divide n  
    while (n % 2 == 0)  
    {  
        //cout << 2 << " ";  
        c++;
        n = n/2;  
    }  
  
    // n must be odd at this point. So we can skip  
    // one element (Note i = i +2)  
    for (ll i = 3; i <= sqrt(n); i = i + 2)  
    {  
        // While i divides n, print i and divide n  
        while (n % i == 0)  
        {  
            //cout << i << " ";  
            c++;
            n = n/i;  
        }  
    }  
  
    // This condition is to handle the case when n  
    // is a prime number greater than 2  
    if (n > 2)  
       // cout << n << " "; 
       c++;
}  
int main()
{
    fio;
    ll n;
    cin>>n;
    //sieve();
    ll i=2;
    ll c=0;
    if(isPrime(n)==true)
    c=1,n=1;
    while(n>1)
    {
        if(n%i==0)
        {
            c++;
            n/=i;
            i++;
        }
        else
        i++;
    }
    /*
    c =primeFactors(n);
    //c=p.size();
    */
    cout<<c<<endl;
}