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
using namespace std;

ll ary[1000005]={0};
void solve1();
void solve()
{
    int t;cin>>t;
    while(t--)
    {
        solve1();
    }
}
int main()
{
    solve1();
    return 0;
}
void solve1()
{
    ll n;
    cin>>n;
    if(n==1)
    {
        cout<<0<<endl;
        return;
    }
    if(n==2)
    {
        cout<<1<<endl;return;
    }
    ll arr[101]={0};
    arr[0]=1;arr[1]=1;
    for(int i=2;i<101;i++)
    {
        arr[i]=arr[i-1]+i;
    }
    ll b=2;
    while(b<=((ll)sqrt(n)))
    {
        if(n%b==0)
        {
            ary[b]++;
            n/=b;
        }
        else
        {
            b++;
        }
    }
    ll ans=0;
    if(n<1000000)
    {ary[n]++;}
    else
    {
        ans++;
    }
    for(int i=2;i<1000005;i++)
    {
        if(ary[i]>0)
        {
            ll j=1;
           // cout<<i<<endl;
            while(ary[i]>=arr[j+1])
            {
                j++;
            }
            ans+=j;
        }
    }
    cout<<ans<<endl;
}