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 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
import java.math.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        long n = sc.nextLong();
        HashMap<Long,Integer> map = getPrimeFactorizationMap(n);
        long ans = 0;
        int cnt = 0;
        int size = map.size();
        for(Map.Entry<Long,Integer> e : map.entrySet()){
            cnt += e.getValue();
        }
        for(int i = 1; i < 100000000; i++){
            for(int j = 0; j < size; j++){
                cnt -= i;
                if(cnt < 0){
                    System.out.println(ans);
                    return;
                }else{
                    ans++;
                }
            }
        }
        out.close();
    }
    
    static HashMap<Long,Integer> getPrimeFactorizationMap(long x) {
        HashMap<Long,Integer> result = new HashMap<Long,Integer>();
        int cnt = 0;
        while (x >= 2L && x % 2L == 0) {
            cnt++;
            x /= 2;
        }
        if(cnt > 0){
            result.put(2L,cnt);
            cnt = 0;
        }
        long d = 3;
        long q = x / d;
        while (q >= d) {
            cnt = 0;
            while (x % d == 0) {
                cnt++;
                x /= d;
            }
            if(cnt > 0){
                result.put(d,cnt);
            }
            d += 2;
            q = x / d;
        }
        if(x != 1) result.put(x,1);
        
        return result;
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;
    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    } 
}