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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;
import java.util.SortedMap;
import java.util.TreeMap;
import java.util.concurrent.atomic.AtomicInteger;

public class Main {

  public static void main(String[] args) {
    resolve(System.in, System.out);
  }

  static void resolve(InputStream is, PrintStream pw) {
    Scanner sc = new Scanner(is);
    long n = Long.parseLong(sc.next());
    SortedMap<Long, AtomicInteger> soinus = new TreeMap<>();
    if (n == 1) {
      pw.println(0);
      return;
    }
    long root = (long) Math.sqrt(n) + 1;
    for (int i = 2; i <= root && i <= n; i++) {
      while (n % i == 0) {
        soinus.computeIfAbsent(
          (long) i,
          k -> new AtomicInteger(0)
        ).incrementAndGet();
        n /= i;
      }
    }
    if (soinus.isEmpty()) {
      soinus.put(n, new AtomicInteger(1));
    }
    SortedMap<Integer, AtomicInteger> eCountMap = new TreeMap<>();
    for (AtomicInteger v : soinus.values()) {
      eCountMap
        .computeIfAbsent(v.intValue(), k -> new AtomicInteger(0))
        .incrementAndGet();
    }
    long ans = 0;
    int totalE = 0;
    for (int i = 1; totalE <= eCountMap.lastKey(); i++) {
      totalE += i;
      long nextTotalE = totalE + i + 1;
      for (int j = totalE; j < nextTotalE; j++) {
        if (eCountMap.containsKey(j)) {
          ans += eCountMap.get(j).intValue() * i;
        }
      }
    }
    pw.println(ans);
  }
}