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<iostream>
#include<vector>
#include<algorithm>
#include <tuple>
#include <utility>
#include<string>
#include <cmath>
#include <queue>
#define ll long long
using namespace std;

double power( double a, ll n ){
  double r;
  ll abs_n;

  r = 1; abs_n = n < 0 ? -n : n;
  while ( abs_n > 0 ) {
    if ( abs_n % 2 == 1 ) { r *= a; }
    a *= a;
    abs_n /= 2;
  }
  return n >= 0 ? r : 1/r;
}

double nthroot( double a, int n ){
  double x, y;

  if ( a <= 0 || n < 0 ) { return 0; }
  if ( n == 0 ) { return 1; }

  /* a > 0 かつ n > 0 の場合 */
  y = (a > 1) ? a : 1;
  do {
    x = y;
    y = ((n-1)*x + a/power(x, n-1)) / n;
  } while ( y < x );

  return y;
}

int is_power(ll a )
{
  ll i; 

  if ( a == 1 ) { return 1; }

  for ( i = 2; i <= floor(log2(a)); i++ ) {
    if ( power(floor(nthroot(a, i)), i) == a ) { 
      return floor(nthroot(a, i));
    }
  }

  return 0;
}

bool is_prime(ll n){
    switch(n){
        case 0: // fall-through
        case 1: return false;
        case 2: // fall-through
        case 3: return true;
    } // n > 3 が保証された

    if(n % 2 == 0 || n % 3 == 0) return false;
    // n は 2 と 3 のいずれの倍数でもないことが保証された
    // これより n は (6の倍数)-1 か (6の倍数)+1 である

    // 6の倍数前後の数を使って試し割りをする
    for(ll i = 5; i * i <= n; i += 6) {
        if(n % i     == 0) return false; // (6の倍数)-1
        if(n % (i+2) == 0) return false; // (6の倍数)+1
    }

    return true;
}

vector<long long> divisor(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    sort(ret.begin(), ret.end()); // 昇順に並べる
    return ret;
}

int main() {
    ll N = 0; cin >> N;
    ll ans=0;
    vector <ll> A = divisor(N);

    for(int j=0; j < A.size(); ++j ){
        int i = A[j];
        if(N%i == 0){
            if(is_prime(i)){
                ans++;
                N = N/i;
            }else if(is_prime(is_power(i))){
                ans++;
                N = N/i;
            }
        }
    }
    cout << ans << endl;
}