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: Wrong Answer)

	#include<bits/stdc++.h>
	#include <ext/pb_ds/assoc_container.hpp> 
	#include <ext/pb_ds/tree_policy.hpp> 
	using namespace __gnu_pbds;
	using namespace std;
	#define lld long long int 
	#define vll vector<long long int>
	#define vs vector<string> 
	#define vint vector<int> 
	#define vp vector<pair<int,int>>
	#define vpll vector<pair<long long int , long long int>>
	#define forl(i,a,b) for(int i = a ; i <= b ; i++)
	#define forln(i,a,b) for(int i = a; i < b ; i++)
	#define forb(i,b,a) for(int i = b ; i >= a ; i--)
	#define forbn(i,b,a) for(int i = b ; i > a ; i--)
	lld mod = 1e9 + 7;
	#define len(str) lld len = str.length();
	#define ps push_back
	#define B begin
	#define E end
	#define mk make_pair
	#define F first 
	#define S second 
	#define FAST ios_base::sync_with_stdio(false),cin.tie(NULL)
	#define test()                  lld t; cin>>t; while(t--)

	typedef tree< pair<lld, lld>, null_type,less<pair<lld, lld>>,rb_tree_tag,
	tree_order_statistics_node_update> ordered_set;

	#define o_set tree<lld, null_type,greater_equal<lld>, rb_tree_tag,tree_order_statistics_node_update> 

	/*
	struct cmp
	{
		bool operator()(const pair<int,int>& l , const pair<int,int>&r){
			if(l.F > r.F) return true;
			if(l.F == r.F){
				if(l.S < r.S) return true;
				else return false;
			}else return false;
		}
	};

	*/
	lld power(lld x, lld y) 
	{ lld temp; if( y == 0) return 1; temp = power((x%mod), (y%mod)/2); if (y%2 == 0) return ((temp%mod)*(temp%mod))%mod;else return (((x%mod)*(temp%mod))%mod*(temp%mod))%mod;} 


	void solution( void );

	int main()
	{
		FAST;
		solution();
		return 0;
	}
	void solution()
	{
		lld n;
		cin>>n;
		if(n == 1){ cout<<"0"<<endl; return ;}
		map<lld,lld>mp;
		lld nn = n;
		for(lld i =2; i <= max((lld)10000100,(lld)sqrt(nn)+1) ; i++){
			while(n%i == 0){
				mp[i]++;
				n /= i;
			}
		}
		lld cnt = 0;
		if(mp.size() == 0){cout<<"1"; return;}
		for(auto it:mp){	
			// cout<<it.S<<endl;
			nn = it.S;
			lld k = 1;
			while(nn > 0){
				// cout<<k<<" ";
				nn -= k;
				if(nn >= 0)
					cnt++;
				k += 1;

			}
		}
		cout<<cnt<<endl;
	}

	// void sol()
	// {
	// 	double a,b;
	// 	cin>>a>>b;
	// 	lld tmp = a*b;
	// 	cout<<tmp;
	// }