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: Runtime Error)

//NAME-ADITYYA OJHA
/*
 *     Heart beats fast
 *         Colors and promises
 *             How to be brave
 *                 How can I love when I am afraid to fall...
                                             (courtesy-amnesiac_dusk)
 */
#include <bits/stdc++.h>
#define ll long long int
#define forin for(int i=0;i<n;i++)
#define forjn for(int j=0;j<n;j++)
#define pb push_back
#define vx vector<ll> x
#define vy  vector<ll> y
#define fas1  ios_base::sync_with_stdio(false)
#define fas2   cin.tie(NULL)
#define fas3   cout.tie(NULL)
#define  f   first
#define  s   second
#define mod 1000000007
 const ll N=1e6+ 100;

using namespace std;

//check for corner cases(for n==1)
// check for int overflow
// run on some sample cases before submitting





   vx;

void primeFactors(ll n)
{

    while (n % 2 == 0)
    {
                 x.pb(2);
        n = n/2;
    }


    for (ll i = 3; i <= sqrt(n); i = i + 2)
    {

        while (n % i == 0)
        {
            x.pb(i);
            n = n/i;
        }
    }


    if (n > 2)
       x.pb(n);
}


int main()
{
    ll n;
    cin>>n;
    primeFactors(n);

    if(x.size()==1){
    cout<<1<<endl;
    return 0;}

    ll freq[N];
    memset(freq,0,sizeof(freq));
    
    


    for(ll i=0;i<x.size();i++)
    {
        freq[x[i]]++;
    }
    ll ans=0;

    for(int i=0;i<N;i++)
    {
         if(freq[i]!=0){
        ll p=1;
        ll sum=0;
        ll cnt=0;
        while(freq[i]>sum)
        {
            sum=(p*(p+1))/2;
           // cout<<sum<<endl;
            p++;
            if(sum>freq[i])break;
            else
            cnt++;
        }
    
         //cout<<freq[i]<<" "<<p<<endl;

        ans+=cnt;
         }

    }
    cout<<ans<<endl;



}