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: Runtime Error)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int N = 1000009, i, j;
vector<int> primes;
void collect_primes(){
	primes.push_back(2);
	primes.push_back(3);
	primes.push_back(5);
	for(i=7; i<=N; i+=2){
		bool is = 1;
		for(j=0; primes[j]*primes[j] <= i; j++){
			if(i % primes[j] == 0){
				is = 0;
				break;
			}
		}if(is){
			primes.push_back(i);
		}	
	}
}
int main(){
	ll n;
	cin >> n;
	if(n == 1){
		cout << 0;
		return 0;
	}
	bool ok = 1;
	for(i=2;i*i<=n;i++){
		if(n%i==0){
			ok = 0;
			break;
		}
	}
	if(ok){
		cout << 1;
		return 0 ;
	}
	
	collect_primes();
	ll ans = 0;
	i = 0;
	ll now = 2;
	while(n>1){
		if(now<=n&&n%now == 0){
			n/=now;
			now = now * primes[i];
			ans++;
		}else if(now>n){
			while(n%primes[i] == 0){
				n/=primes[i];
			}now = primes[i+1];
			i++;
		}
		else {
			i++;
			now = primes[i];
		}
	}
	cout << ans;
}