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<bits/stdc++.h>
using namespace std; 
bool ispower(int x,set<int> s)
{
	int f=0;
  int p=1;
	while(x%2==0)
	{
      p*=2;
		x=x/2;
		f=1;
	}
	if(f==1 && x!=1)return false;
	if(x==1 && s.find(p)==s.end())return true;
	int t=sqrt(x);
	for(int i=3;i<=t;i+=2)
	{
      int p=1;
		while(x%i==0)
		{
			f=1;
			x=x/i;
          p*=i;
		}
		if(f==1 && (x!=1))return false;
		if(f==1 && (x==1 && s.find(p)==s.end()))return true;
	}
	return false;
}
bool isprime(int x)
{
	for(int i=2;i<=sqrt(x);i++)if(x%i==0)return false;
	return true;
}
int main()
{
	
	long long x;
	cin>>x;
	long long ans=0;
	int r=sqrt(x);
	set<int> s;
	bool seive[r+1];
	for(int i=2;i<=r;i++)seive[i]=true;
	for(int i=2;i<=r;i++)
	{
		if(seive[i])
		{
			for(int j=i;j<=r;j*=i)
			{
				if(x%j==0)
				{
					x=x/j;
					ans++;
					s.insert(j);
				}
			}
			for(int j=2*i;j<=r;j+=i)
			{
				seive[j]=false;
			}
		}
	}
	if(x>=2 && s.find(x)==s.end() && (ispower(x,s) || isprime(x)))ans++;
	cout<<ans;
	return 0;
}