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<queue>
#include<algorithm>
#include<vector>
#include<set>
#include<string.h>
#include<math.h>
#define ll long long
using namespace std;
const int MAX = 1000000;
int cnt[MAX];
int main() {
	ll n;
	cin >> n;
	int i;
	ll k=n;
	bool f=1;
	int cnt = 0;
	for (i = 2; i <= sqrt(k)&&n>=i; i++) {
		f = 1;
		for (int j = 2; j <= sqrt(i); j++) {
			if (!i % j) {
				f = 0;
				break;
			}
		}
		if (f) {
			ll j = i;
			while (!(n % j)&&n>=j)
			{
				n /= j;
				cnt++;
				j *= (ll)i;
			}
		}
	}
	if (n==k&&cnt==0) {
		cout << 1;
		return 0;
	}
	cout << cnt;
	return 0;
}