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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define ll int64_t
#define _GLIBCXX_DEBUG
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
//約数列挙(√N)
vector< int64_t > divisor(int64_t n) {
  vector< int64_t > ret;
  for(int64_t i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}
//素因数分解(√N)
map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
//素数テーブル(NloglogN)
vector< bool > prime_table(int n) {
  vector< bool > prime(n + 1, true);
  if(n >= 0) prime[0] = false;
  if(n >= 1) prime[1] = false;
  for(int i = 2; i * i <= n; i++) {
    if(!prime[i]) continue;
    for(int j = i + i; j <= n; j += i) {
      prime[j] = false;
    }
  }
  return prime;
}
//二項係数(K)
template< typename T >
T binomial(int64_t N, int64_t K) {
  if(K < 0 || N < K) return 0;
  T ret = 1;
  for(T i = 1; i <= K; ++i) {
    ret *= N--;
    ret /= i;
  }
  return ret;
}
//二項係数テーブル(N^2)
template< typename T >
vector< vector< T > > binomial_table(int N) {
  vector< vector< T > > mat(N + 1, vector< T >(N + 1));
  for(int i = 0; i <= N; i++) {
    for(int j = 0; j <= i; j++) {
      if(j == 0 || j == i) mat[i][j] = 1;
      else mat[i][j] = mat[i - 1][j - 1] + mat[i - 1][j];
    }
  }
  return mat;
}

int main(){
  ll n;
  cin >> n;
  int ans=0;
  map m=prime_factor(n);
  for(auto i:m){
    int c=i.second;
    int b=1;
    while(c>=b){
      c-=b;
      b++;
      ans++;
    }
  }
  cout <<ans <<endl;
}