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>
#include<iostream>
#include<string>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<iomanip>
#include<cstdlib>
#include<ctime>
#include<set>
#include<map>
#include<utility>
#include<queue>
#include<vector>
#include<stack>
#include<sstream>
#include<algorithm>
/************************************************/
#define rep(i,n) for(int i=0;i<n;i++)
#define m_p make_pair
#define pb push_back
#define fr first
#define se second
#define forn(i,a,n) for(int i=a;i<n;i++)
#define foreach(i,c) for(__typeof(c.begin())i=(c.begin());i!=(c).end();i++)
#define pii pair<int,int>
#define vi vector<int>
#define ll long long
#define sz(s) s.size()
#define all(s) s.begin(),s.end()
#define zero(x) memset(x,0,sizeof(x))
#define vii vector<pair<int,int> >
#define mpis map<int,string>
#define mpii map<int,int>
#define mpsi map<string,int>
#define re return
#define mod 1000000007
/************************************************/
using namespace std;

long long get(){
	char c=getchar();
	long long x=0LL;
	while(c<'0'||c>'9')
	c=getchar();
	while(c>='0'&&c<='9'){
		x*=10LL;
		x+=(c-'0');
		c=getchar();
	}
	return x;
}
int ans;
bool isprime(ll x){
	for(ll i=2;i*i<=x;i++){
		if(x%i==0)
		re false;
	}
	re true;
}
bool ip[1000001]; 
vector<ll> prime;
vi j;
vi v;
int main(){ 
	ios::sync_with_stdio(0);
	ll sum=0;
	rep(i,100){
		sum+=i+1;
		j.pb(sum);
	}
	ll n;
	cin>>n;
	if(n==1){
		cout<<0;
		re 0;
	}
	if(isprime(n)){
		cout<<1;
		re 0;
	}
	for(ll i=2;i<=1000000;i++){
		if(ip[i])
		continue;
		for(ll j=i*i;j<=1000000;j+=i)
		ip[j]=true; 
	}
	for(int i=2;i<=1000000;i++)
	if(!ip[i])
	prime.pb(i);
	int now=0;
	while(n>1){
		int nn=0;
		while(now<(int)(sz(prime))&&n%prime[now])now++;
		if(now==(int)(sz(prime))){
			ans++;
			break;
		}
		while(n%prime[now]==0){
			nn++;
			n/=prime[now];
		}
		v.pb(nn);
	}
	rep(i,sz(v)){
		int pos=upper_bound(j.begin(),j.end(),v[i])-j.begin();
		ans+=pos;
	}
	cout<<ans;
	re 0;
}