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.

Examples

Input

24


Output

3


Input

1


Output

0


Input

64


Output

3


Input

1000000007


Output

1


Input

997764507000


Output

7
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);
        ArrayList<Integer> li = new ArrayList<Integer>();
        for(Map.Entry<Long,Integer> e : map.entrySet()){
            li.add(e.getValue());
        }
        long ans = 0;
        Integer[] arr = li.toArray(new Integer[li.size()]);
        for(int i = 0; i < arr.length; i++){
            for(int j = 1 ;; j++){
                if(arr[i] >= j){
                    arr[i] -= j;
                    ans++;
                }else{
                    break;
                }
            }
        }
        out.println(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,result.getOrDefault(x,0)+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;
    } 
}
