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 <bits/stdc++.h>
using namespace std;

bool prime(long long x) {
	if (x <= 1LL) return false;
  	if (x == 2LL) return true;
  	for (long long i = 2LL; i * i <= x; ++i) {
    	if (x % i == 0LL) return false;  
    }
  	return true;
}

int main() {
	long long n; cin >> n;
  	map<long long, long long> m;
  	for (long long i = 2LL; i * i <= n; ++i) {
      	if (prime(i)) {
        	long long nn = n;
          	while (nn % i == 0LL) {
            	++m[i], nn /= i;  
            }
        }
    }
  	set<long long> s;
    for (auto tr : m) {
      	//cout << tr.first << " " << tr.second << endl;
    	long long cr = tr.second, mi = 1LL, po = tr.first;
      	while (cr >= mi) {
        	s.insert(po);
          	cr -= mi++;
          	po *= tr.first;
        }
    }
  	long long ans = 0LL;
	for (long long x : s) {
      	//cout << x << endl;
    	if (n % x == 0LL) {
        	++ans, n /= x;  
        }
    }
  	cout << ans << endl;
  	return 0;
}