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 <iostream>
#include <vector>
using namespace std;

int main()
{
	auto p = vector<bool>(1000000);
	auto pl = vector<int>();
	p[0] = p[1] = true;
	for(int i = 2; i < 1000000; ++i){
		if(p[i]) continue;
		pl.push_back(i);
		for(int j = i + i; j < 1000000; j += i){
			p[j] = true;
		}
	}
	long long int n;
	cin >> n;
	int ans = 0;
	for(int i = 0; i < (int)pl.size(); ++i){
		if(n == 1) break;
		int cnt = 0;
		while(n % pl[i] == 0){
			++cnt;
			n /= pl[i];
		}
		for(int j = 1;; ++j){
			cnt -= j;
			if(cnt < 0) break;
			++ans;
		}
	}
	if(n > 1) ++ans;
	cout << ans << endl;
	return 0;
}