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<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
long long zz[1000]={0};
bool isPrime(long long x){
	if(x==1)return 0;
	for(long long i=2;i<=sqrt(x);i++){
		if(x%i==0)return 0;
	}
	return 1;
}
bool notExist(int x){
	int i;
	for(i=1;zz[i]!=0;i++){
		if(x==zz[i])
			return 0;
	}
	zz[i]=x;
	return 1;
}
int main()
{
	long long n,p,ans=0,e,z;
	cin >> n;
	if(isPrime(n)==1){
		cout << 1 << endl;
	}
	else{
		for(p=2;p<=n;p++){
			if(isPrime(p)==0)
				continue;
			for(e=1;pow(p,e)<=n;e++){
				z=pow(p,e);
				if(n%z!=0){
					if(e==1)break;
					else{
						z=sqrt(z);
						if(n%z==0){
							n=n/z;
						}
					}
				}
				if(notExist(z)==1){
					ans++;
					n=n/z;
				}
			}
			if(isPrime(n)==1){
				ans++;
				break;
			}
		}
		cout << ans << endl;
	}
	return 0;
}