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>
using namespace std;
typedef long long LL;
const LL MOD = 1000000007;
#define lcm(x, y) ((x) * (y) / __gcd((x), (y)))
#define pb push_back
#define all(c) (c).begin(), (c).end()
#define debug(x) cout << #x << " : " << x << endl
#define part cout << "----------------------------------\n";
#define fastinput                     \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
LL cal(LL n)
{
    return (n*(n+1))/2;
}

int main()
{
    fastinput;
    LL n,i,j,k,t,temp;
    cin>>n;
    map<LL,LL> freq;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            while(n%i==0)
            {
                freq[i]++;
                n/=i;
            }
        }
    }

    if(n!=1)
    {
        freq[n]++;
    }

    LL ans=0;

    for(auto x:freq)
    {
        LL f=x.second;
       // debug(x.first);
        //debug(x.second);
        LL lb=0,ub=50;
        LL pos=-1,mid=-1;
        while(lb<=ub)
        {
            mid=(lb+ub)/2;
            if(cal(mid)>f)
            {
                pos=mid;
                ub=mid-1;
            }
            else
            {
                lb=mid+1;
            }            
        }

        pos--;
        //debug(pos);
       // part;
        ans+=pos;

    }
   // debug(ans);
   cout<<ans<<endl;

    return 0;
}