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: Runtime Error)

#include <stack>
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <array>
#include <iomanip>
#include <string>
using namespace std;
int gcd(int a, int b)
{
   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }
}
int mod = 1000000000 + 7;
vector<int> v;
class Calc
{
    // 宣言
    int a;

    public:
        // 素因数分解
        void decompositPrime(long long a);
};

/*
 * 素因数分解
 */
void Calc::decompositPrime(long long n)
{
    // 割る数の初期値
    a = 2;
    // √n ≧ a ( n ≧ a * a ) の間ループ処理
    while (n >= a * a) {
        // a で割り切れたら、a は素因数
        // そして、割られる数を a で割る
        // a で割り切れなかったら、 a を 1 増加させる
        if (n % a == 0) {
            //printf("%d * ", a);
            v.push_back(a);
            n /= a;
        } else {
            a++;
        }
    }
    // 最後に残った n は素因数
    v.push_back(n);
    //printf("%d\n", n);
}

int main(){
    //std::stack<int> stack;
    long long n;
    cin >> n;
    int sq = (int)sqrt(n);
    vector<long long> cnt(sq+1);
    long long ans = 0;
    if(n == 1){
        ans = 0;
    }else{
        Calc objCalc;
        objCalc.decompositPrime(n);
        if(v.size() == 1){
            ans = 1;
            cout << ans << endl;
            return 0;
        }
        for (auto &num:v){
            cnt[num] ++;
        }
        int tmp;
        for(int i = 2; i <= sq; i++){
            //cout << cnt[i] << endl;
            tmp = 1;
            while(tmp <= cnt[i]){
                ans ++;
                cnt[i] -= tmp;
                tmp ++;
            }
        }
        
    }
    cout << ans << endl;
    return 0;
}