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;

#define ll long long
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define x first
#define y second
const int INF=1e6+1;

bool isp[INF];

int main(){
	ll n;
	cin>>n;
	vector<ll> pr;
	for (int i=2; i<INF; i++){
		if (!isp[i]) pr.pb(i);
		for (int j=i+i; j<INF; j+=i)
			isp[j]=1;
	}
	ll ans=0;
	ll t=n;
	for (int i=0; i<pr.size(); i++){
		ll cnt=0;
		while(t%pr[i]==0){
			t/=pr[i];
			cnt++;
		}
		ll cur=1,ret=0;
		while (cnt>=cur){
			cnt-=cur;
			ret++;
			cur++;
		}
		ans+=ret;
	}
	if (t>1) ans++;
	cout<<ans<<endl;
	return 0;
}