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 ll long long
#define IO std::ios::sync_with_stdio(false);std::cin.tie(0)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define mset(s,_) memset(s,_,sizeof s)
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define mk make_pair
using namespace std;
const int maxn=1000010;
ll vp[maxn+10],p[maxn+10];
void prime(){
	  vp[0]=vp[1] = 1;    //0和1为合数
	for(ll i = 2; i < maxn; i++){
		if(!vp[i]) p[++p[0]] = i;        //p[]里面的是素数
		for(ll j = 1; j <=p[0] && i * p[j]<maxn; j++){
			vp[i * p[j]] = 1;
			if(i % p[j] == 0) break;
		}
	}
}
vector<ll>v;
void init(){
	for(ll i=1;i<=p[0];i++){
		for(ll j=p[i];j<=1000000000000;j*=p[i]){
			v.push_back(j);
		}
	}
}
int main(){
 prime();
 init();
 ll n;
 sort(v.begin(),v.end());
 cin>>n;
 int ans=0,w1=0;
 for(int i=0;i<v.size();i++){
 	if(n%v[i]==0) n/=v[i],ans++,vp[v[i]]=1;
	if(n<v[i]) break;
  }
	if(n==1) return cout<<ans<<'\n',0;
	else{
		for(int i=2;i<=1000000;i++){
			if(!vp[i]&&n%i==0){
				for(ll j=i;j<=n;j*=i){
					if(j==n) {ans++;return cout<<ans<<'\n',0;  }
				}
			}
		}
	}
	cout<<ans+1<<'\n';
}