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 int
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define mod 1000000007
#define PI 3.141592653589793238
#define all(a) a.begin(),a.end()
#define dbg(n) cout<<#n<<' '<<n<<endl;
#define dbg_v(v) cout<<#v<<":";for(auto x:v) cout<<" "<<x; cout<<endl;
int power(ll x,ll y){ll res = 1;while(y>0){if(y&1)res = (res*x)%mod;y=y>>1;x=(x*x)%mod;}return res;}
int gcd(int a,int b){
	if(!a)
		return b;
	return gcd(b%a,a);
}
int main(){ 
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);	
	ll n;
	cin>>n;
	ll k=n;
	vector<int>a;
	ll sum[1000001];
	sum[0]=0;
	sum[1]=1;
	for(int i=2;i<1000001;i++)
		sum[i]=sum[i-1]+i;
	for(ll i=2;i*i<=n;i++){
		int ct=0;
		while(k%i==0){
			k/=i;
			ct++;
		}
		if(ct){
		a.pb(ct);
		}
	}
	if(k>1)
		a.pb(1);
	ll ans=0;
	for(int i=0;i<(int)a.size();i++){
		auto l=lower_bound(sum,sum+a[i],a[i]);
		if(sum[l-sum]>a[i])
			ans+=(ll)(l-sum)-1;
		else
			ans+=(ll)(l-sum);
	}
	cout<<ans<<endl;
}