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>
#define ss second
#define ff first
#define all(x) x.begin(), x.end()
 
using namespace std;
using ll = long long;
using pii = pair<int, int>;
 
const int oo = 1e9 + 7;
const int mod = 1e9 + 7;
const double PI = acos(-1);

bool isprime[1000100]={};
vector<int> primos;

void sieve(){
    isprime[0] = isprime[1] = true;
    primos.push_back(2);
    for (int i=3; i<1010; i+=2){
        if (!isprime[i]){
            primos.push_back(i);
            for (int j=i*i, k = (i << 1); j<=1000100; j+= k){
                isprime[j] = true;
            }
        }
    }

    for (int i=1011; i<1000100; i+=2){
        if (!isprime[i]) primos.push_back(i);
    }
}
 
 
int main (){
    sieve();
    ll n, ans = 0, power;
    cin >> n;

    for (auto p: primos){
        if (p > n) break;
        power = 0;
        while (n%p == 0) n/=p, power++;
        if (power) ans += (1 + ll(sqrt(1 + 8*power)))/2 - 1;
    }

    if (n > 1) ans++;
    cout << ans  << endl;

    return 0;
}