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 watch(x) cout << (#x) << " is " << (x) << endl
const int mod = 1e9 + 7;

using namespace std;

set<long long> factorization;
map<long long, int> app;

void factoTrial(long long n) 
{
    while (n % 2 == 0) 
    {
        factorization.insert(2);
        app[2]++;
        n /= 2;
    }
    for (long long d = 3; d * d <= n; d += 2) 
    {
        while (n % d == 0) 
        {
            factorization.insert(d);
            app[d]++;
            n /= d;
        }
    }
    if (n > 1)
    {
        factorization.insert(n);
        app[n]++;
    }

}

int main()
{
	#ifdef ME
		freopen("input.txt", "r", stdin);
		freopen("output.txt", "w", stdout);
	#endif

	long long n, ans = 0; cin >> n;
	factoTrial(n);
	for (auto &x : factorization)
	{
		//watch(x);
		for(int j = 1; j <= app[x]; j++)
		{
			//watch(app[x]);
			app[x] -= j;
			ans++;
		}
	}

	cout << ans << endl;
	return 0;
}