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)

#include<bits/stdc++.h>
#define ll long long 
//#define inf 1000000000000000000
using namespace std;
ll mod=1000000000000000001;
vector<ll> primeFactors(ll n)  
{  
    vector<ll> fact;
    // Print the number of 2s that divide n  
    while (n % 2 == 0)  
    {  
        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 (ll i = 3; i <= sqrt(n); i = i + 2)  
    {  
        // While i divides n, print i and divide n  
        while (n % i == 0)  
        {  
            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);
    return fact;
}  

int main(){
    ll n;
    cin>>n;
    vector<ll>fac=primeFactors(n);
    ll ans=0;
    for(int i=0; i<fac.size(); i++){
        int j=i;
        ll count=0;
        while(j<fac.size() && fac[i]==fac[j]){
            count++;
            j++;
        }
        ll y=1;
        y+=8*count;
        y=sqrt(y);
        y-=1;
        y/=2;
        ans+=y;
        i=j-1;
    }
    cout<<ans;
    return 0;
}