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>
using namespace std;
#define int long long int
#define double long double
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mod 1000000007
#define endl "\n"
#define F first
#define S second
#define pb push_back
#define loop(a,b) for(int i=a; i<b; i++)
#define loopr(a,b) for(int i=a;i>=b;i--)
int getval(int x)
{
     int p=sqrt(8*x+1)-1;
     int ans=p/2;
     return ans;
     /*if(ans<=2*x)return p+1;
     else return p;*/
}
void solve()
{
	int n;cin>>n;
	int e=sqrt(n);
	int count=0;
	int temp=0;
	while(n%2==0)
	{
	     n/=2;
	     temp++;
	}
//	cout<<temp<<"\n";
	count+=getval(temp);
	for(int i=3;i<=e && n!=1;i+=2)
	{
	     temp=0;
	     while(n%i==0)
	     {
	          n/=i;
	          temp++;
	     }
	     if(temp!=0){
	     count+=getval(temp);
	     //cout<<i<<" "<<temp<<" "<<getval(temp)<<"\n";
	     }
	}
	if(n!=1)cout<<1<<"\n";
	else
	cout<<count;
	return;
}
int32_t main() 
{ 
	fast;
	int t=1;
	//cin>>t;
	while(t--)
	{
		solve();
	}
	return 0; 
}