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 rep(i,x,y) for(int i=x;i<y;i++)
#define rept(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=y;i>=x;i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define pii pair<int,int>
//	ios::sync_with_stdio(false);
//	cin.tie(0);
const int maxx = 1e7+7;
const int maxn = 1e9+7;
typedef long long ll;
using namespace std;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
ll notprime[maxx]={1,1};
ll prime[maxx];
ll cnt[maxx];
ll num,n,ans;
void Pri(ll n){
	for(ll i=2;i<maxx;i++){
		if(!notprime[i])prime[num++]=i;
		for(ll j=0;j<num && i*prime[j]<maxx;j++){
			notprime[i*prime[j]]=1;
			if(!(i%prime[j]))break;
		} 
	}
}
int sushu(ll a){
	if(a==1)return 0;
	else if(a==2||a==3||a==5){
		return 1;
	}else{
		for(int i=2;i<=sqrt(a);i++){
			if(a%i==0){
				return 0;
			}
		}
		return 1;
	}
}
int main(){
		ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	Pri(10000007);
	if(sushu(n)){
		cout<<1<<endl;
	//	cout<<"K"<<endl;
		return 0;
	}
	rep(i,0,num){
		if(n==1)break;
		int op=0;
		while(n%prime[i]==0)op++,n/=prime[i];
		if(op<3&&op>=1)ans+=1;
		else if(op>=3&&op<6)ans+=2;
		else if(op>=6&&op<10)ans+=3;
		else if(op>=10&&op<15)ans+=4;
		else if(op>=15&&op<21)ans+=5;
		else if(op>=21&&op<28)ans+=6;
		else if(op>=28&&op<36)ans+=7;
		else if(op>=36&&op<45)ans+=8;
		else if(op>=45&&op<55)ans+=9;
	}
	cout<<ans<<endl;
	return 0;
}