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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
typedef long long ll;

int dandan(ll n)
{
    int ret=0;
    if(n==0)
    {
        ret=0;
    }
    else
    {
        int i=1;
        while (n>=i)
        {
            ret=ret+1;
            n=n-i;
            i=i+1;
        }
        return ret;
    }
    
    
}

int main()
{
    ll n;
    scanf("%lld",&n);

    if(n==1)
    {
        printf("0\n");
    }

    else
    {
    

    ll m=n;
    ll p[100];
    for(int a=0;a<100;a++)
    {
        p[a]=1;
    }

    ll i=2;
    ll cp=0;

    while(m>1&&i<=sqrt(n))
    {
        if(m%i==0)
        {
            m=m/i;
            p[cp]=i;
            cp=cp+1;
        }
        else
        {
            i=i+1;
        }
    }



    cp=0;
    ll q[100];
    ll cq=0;
    for(int a=0;a<100;a++)
    {
        q[a]=0;
    }

    q[0]=1;

    while(p[cp+1]!=1)
    {
        if(p[cp]==p[cp+1])
        {
            q[cq]=q[cq]+1;
            cp=cp+1;
        }
        else
        {
            cq=cq+1;
            q[cq]=1;
            cp=cp+1;
        }
    }


    ll ans=0;
    for(int a=0;a<100;a++)
    {
        ans=ans+dandan(q[a]);
    }
    
    printf("%lld\n",ans);


    }

    return 0;
}