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;
using ll = long long;
template<typename T> using v2 = vector<vector<T>>;
template<typename T> inline v2<T> fill(int r, int c, T t){return v2<T>(r, vector<T>(c, t));}

int bs(int sum){
    int lo = 0;
    int hi = sum;
    int ans = -1;
    while(lo <= hi){
        int mid = (lo+hi)/2;
        if(mid*(mid+1) <= 2*sum){
            ans = mid;
            lo = mid+1;
        }
        else{
            hi = mid-1;
        }
    }
    return ans;
}
ll n;
void solve(){
    cin >> n;
    ll s = sqrt(n);
    int ans = 0;
    for(int i = 2; i <= s; i++){
        int times = 0;
        while(n%i == 0){
            n /= i;
            times++;
        }
        //cout << i << " " << times << " " << bs(times) << '\n';
        ans += bs(times);
    }
    cout << (!ans ? 1 : ans) << '\n';
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}