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 L 1000000000000000000
#define ll long long int
#define ld long double
using namespace std;
int main(void){
	ll n,i;
	cin>>n;
	if(n==1)
	cout<<0<<endl;
	else{
		ll ans=0,lans;
		ld k;
		set<ll>s;
		map<ll,int>mp;
		while(n%2==0){
			mp[2]++;
			s.insert(2);
			n/=2;
		}
		k=floor((pow(1.0+8.0*mp[2],0.5)-1.0)/2.0);
		lans=(ll)k;
		ans+=k;
		//cout<<mp[2]<<endl;
		for(i=3;i*i<=n;i+=2){
			while (n % i == 0)  
			{  
				mp[i]++;  
				n = n/i;
				s.insert(i);  
			}
		//	cout<<i<<" "<<mp[i]<<endl;
			k=floor((pow(1.0+8.0*mp[i],0.5)-1.0)/2.0);
			lans=(ll)k;
			ans+=k;  
		}
		if (n > 2){  
			s.insert(n);
			mp[n]++;
		}
		k=floor((pow(1.0+8.0*mp[n],0.5)-1.0)/2.0);
			lans=(ll)k;
			ans+=k;  
		//ans+=s.size();
		cout<<ans<<endl;
	}
}