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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=1e6+50;
int prime[maxn];
bool sf[maxn];
int cnt[maxn];
int num=0;
void sushu(){
    memset(sf,true,sizeof(sf));
    for(int i=2;i<=maxn;i++){
        if(sf[i]) prime[++num]=i;
        for(int j=1;j<=num;j++){
            if(i*prime[j]>maxn) break;
            sf[i*prime[j]]=false;
            if(i%prime[j]==0) break;
        }
    }
    sf[1]=false;
    sf[0]=false;  //1 0 特判
}
int main()
{
    memset(cnt,0,sizeof cnt);
    sushu();
    ll n,x;
    scanf("%lld",&n);
    x=sqrt(n);
    ll res=0;
    for(int i=1;prime[i]<x&&i<=n;i++){
        if(n%prime[i]==0){
            while(n%prime[i]==0){
                cnt[i]++;
                n/=prime[i];
            }
            int j=1;
            while(cnt[i]>=j){
                cnt[i]-=j++;
                res++;
            }
        }
    }
    if(n!=1)
        res++;
    printf("%lld\n",res);
    return 0;
}