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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <fstream>
#include <set>
#include <stack>
#include <queue>
#include <list>
using namespace std;
typedef long long int64;
typedef vector<int> vec;
typedef vector<int64> vec64;
#define db(x) cout << "> " << #x << ": " << x << "\n";
#define ss cout << " ";
#define ct(x) cout << x;
#define nn cout << "\n";
#define cts(x) cout << x << " ";
#define ctn(x) cout << x << "\n";
// // // // // // // // // // // // // // // // // // // // // // 
/*                  TEMPLATE - VANILLA                         */
// // // // // // // // // // // // // // // // // // // // // //
int64 q, n;
const int maxn = 1e6+5;
int ciur[maxn + 10];

void solve(){

    return;
}
bool isPrime(int64 n){
    if (n == 1){
        return 0;
    }
    for (int64 i = 2 ; i*i <= n; i++){
        if (n % i == 0){
            return 0;
        }
    }
    return 1;
}

int main(){
    int64 n;
    cin >> n;
    int64 rs =0;
    if (isPrime(n)){
        rs++;
    }
    for (int i = 2; i < maxn; i++){
        if (!ciur[i]){
            for (int j = i+i; j <maxn; j+=i){
                ciur[j] = 1;
            }
        }
    }
    map <int, int> mp;
    int64 q = n;
    for (int i = 2; i <= sqrt(q); i++){
        // cts(i); ctn(ciur[i]); 
        if (!ciur[i] && n % i == 0){
            while (n % i == 0){
                n/=i;
                mp[i]++;
            }
        }
    }
    int64 sofar = 0;
    for (auto i: mp){
        // cts(i.first);
        // ctn(i.second);
        int x = i.second;
        sofar = 1;
        while (x >= sofar){
            rs++;
            x--;
            sofar+=1;
        }
    }
    ctn(rs);
    return 0;
}