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: Time Limit Exceeded)

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<math.h>
#include<map>
#include<iomanip>
#include<queue>

const long long INF = 1e17+7;
const long long MOD = 1e9+7;
const double PI=acos(-1);

using namespace std;

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

int main(){
	
	long long n;
	cin >> n;
	
	if(is_prime(n)==true){
		cout << "1" << endl;
		return 0;
	}
	
	map<long long,int> mp;
	long long num=n;
	for(long long i=2;i<=n;i++){
		while(1){
			if(num%i!=0)break;
			mp[i]++;
			num/=i;
		}
		if(num==1 || num==0)break;
	}
		
	long long cnt=0;
	for(auto itr=mp.begin();itr!=mp.end();itr++){
		int now=itr->second;
		//cout << itr->first << " : " << itr->second << endl;
		for(long long i=1;i<=itr->second;i++){
			now-=i;
			cnt++;
			if(now<0){
				cnt--;
				break;
			}
			if(now==0){
				break;
			}
		}
	}
	cout << cnt << endl;
	
	return 0;
}