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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
#define rep2(i,x,n) for(int i=x; i<(n); i++)
#define ALL(n) begin(n),end(n)
using namespace std;
using P = pair<int, int>;
using ll = long long;

int main()
{
    ll n;
    cin >> n;

    if(n==1){
        cout << 0 << endl;
        exit(0);
    }
    vector<ll> p;
    ll a = 2;
    while (n >= a * a) {
        if (n % a == 0) {
            p.emplace_back(a);
            n /= a;
        } else {
            a++;
        }
    }
    p.emplace_back(n);
    ll ans = 0;
    rep2(i,2,p[p.size()-1]+1){
        int cnt = count(ALL(p),i);
        if(cnt==0) continue;
        if(cnt>=1)ans++;
        if(cnt>=3)ans++;
        if(cnt>=6)ans++;
        if(cnt>=10)ans++;
        if(cnt>=15)ans++;
        if(cnt>=21)ans++;
        if(cnt>=28)ans++;
        if(cnt>=36)ans++;
        if(cnt>=45)ans++;
    }
    cout << ans << endl;
    return 0;
}