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>
#define con continue
#define pb push_back
#define fi first
#define se second
#define fr front()
#define one 1.0
#define inf 1000000000000000000
#define ret return
#define all(vl) vl.begin(),vl.end()
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef map<ll,ll>::iterator itll;
typedef long double ld;
typedef map<ll,ll> mapll;
bool zhi[1000001];
int main()
{
	zhi[0] = zhi[1] = 1;
	vector<ll> vl;
	ll n;cin>>n;
	for(ll i = 2;i<1000001;i++)
	{
		if(!zhi[i])
			vl.pb(i);
		for(ll j = i*i;j<1000001;j+=i)
			zhi[j] = 1;
	}
	ll cnt = 0;
	ll cop = n;
	bool isok = false;
	int sz = vl.size();
	for(int i=0;i<sz;i++)
	{
		if(n%vl[i] == 0)
			isok = true;
	}
	if(n>1e6&&!isok)
	{
		cout<<"1\n";
		return 0;
	}
	for(int i=0;i<sz;i++)
	{
		ll cc = 0;
		while(cop%vl[i] == 0)
		{
			cc++;
			cop/=vl[i];
		}
		ll cn,ccn;
		cn = 1,ccn = 0;
		while(cc>=cn)
		{
			cc-=cn;
			ccn++;
			cn++;
		}
		cnt+=ccn;
	}
	cout<<cnt<<endl;
	return 0;
}