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 rep(i,n) for(int i=0;i<(n);++i)
#define FOR(i,n,j) for(ll i=(j);i<(n);++i)
#define rrep(i,n) for(int i=(n);i>=0;--i)
#define ssort(n) sort((n).begin(),(n).end())
using ll=long long;
using ld=long double;
typedef pair<int,int> P;
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define INF 10101010

int main()
{
    ll n;
    cin >> n;
    int t;
    ll n_tmp=n;
    int count;
    int j;
    ll ans=0;
    FOR(i,sqrtl(n)+1,2){
        count=0;
        while(n_tmp%i==0){
            n_tmp/=i;
            count++;
        }
        j=1;
        while(count>=j){
            count-=j;
            j++;
            ans++;
        }
    }
    if(n_tmp==n&&n!=1){
        ans=1;
    }else if(n==1){
        ans=0;
    }
    n=n_tmp;
    FOR(i,sqrtl(n)+1,2){
        while(n_tmp%i==0){
            n_tmp/=i;
        }
    }
    if(n==n_tmp){
        ans++;
    }
    cout << ans << endl;

}