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;
typedef long long ll;
long long N, ans[1000000], n,d = 0;
int main(){
	//freopen("abc169.inp", "r", stdin);
	//freopen("abc169.out", "w", stdout);
	cin>>n;
	ll cc = sqrt(n);
	for (int i = 2; i <= cc; i++){
		if (n % i == 0){
			++d;
			while (n % i == 0){
				ans[d]++; n = n / i;
			}
		}
	}
	ll _ans = 0;
	//for (int i = 1; i <=d ; i++) cout<<ans[i]<<"\n";
	for (int i = 1; i <= d; i++){
		ll lo = 1, hi = 1e9, res = 0;
		while (lo <= hi){
			ll mid = (lo + hi)/2;
			if (mid * (mid + 1) <= ans[i] * 2){
				res = mid;
				lo = mid + 1;
			}
			else hi = mid - 1;
		} 
		_ans += res;
	}
	cout<<_ans;
}