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)

/*   ꧁༒☬महाकाल☬༒꧂ ꧁༒ঔमहाकालঔ༒꧂
                   
                   ঔৣ۝महाकालᴳᵒᵈ۝ঔৣ            
 
 */ 
#include<bits/stdc++.h>
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template <typename T, typename cmp=less<T>> using oset =tree<T, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
#define ll long long 
#define mod 1000000007
#define inf 100000000000000000
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define _it iterator
#define all(_x) _x.begin(),_x.end()
#define f first
#define s second
#define pb push_back
void mahakal(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	freopen("error.txt","w",stderr);
	#endif
}
const ll N = 1e6+1;
ll solve(ll n){
	ll l=0 , r=1e7;
	ll ans=0;
	while(l<=r){
		ll mid=(l+r)/2;
		ll pj=(mid*(mid+1))/2;
		if(pj<=n){
			ans=mid;
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	}
	return ans ;
}
void solve(){
	ll n ;
	cin>> n ;
	ll ans=0;
	for(ll i=2;i*i<=n ; i++){
		ll ps=0;
		while(n%i==0){
			n/=i;
			ps++;
		}
		ans+=solve(ps);
		// cerr<<ans<<" "<<ps<<endl;
	}
	if(n>1)ans++;
	cout<<ans<<endl;
}
int main(){
	//mahakal(),fast;
	ll t=1 ;
	// cin>> t;
	while(t--)solve();

	return 0;
}