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>
using namespace std;
using ll = long long;
const int MOD = 1e9+7;

bool hantei(long long x){
	for(long long i=2; i*i<=x; i++){
		if(x%i==0) return 0;
	}
	return x>1;
}

int main(){
	ll n; cin>>n;
	map<ll,ll> m;
	if(hantei(n)){
		cout << 1 << endl;
		return 0;
	}
	else if(n==1){
		cout << 0 << endl;
		return 0;
	}
	else{
		ll b{2};
		while(1){
			if(n%b==0){
				m[b]++;
				n/=b;
				if(n==1) break;
				if(hantei(n)){
					m[n]++;
					break;
				}
			}
			else b++;
		}
	}
	vector<ll> a(10);
	for(int i=1; i<10; i++){
		a.at(i)=a.at(i-1)+i;
	}
	ll ans{};
	for(auto i:m){
		ll e = i.second;
		auto it = upper_bound(a.begin(),a.end(),e);
		ans += it - a.begin()-1;
	}
	cout << ans << endl;
	
}