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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.math.*;

public class Main {

    private static Scanner sc;

    public static void main(final String[] args) {
        final Main instance = new Main();
        sc = instance.new Scanner();
        instance.solve();
    }

    private class Scanner {
        String[] s;
        int i;
        BufferedReader br;
        String regex = " ";

        public Scanner() {
            s = new String[0];
            i = 0;
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        @Override
        protected void finalize() throws Throwable {
            try {
                super.finalize();
            } finally {
                destruction();
            }
        }

        private void destruction() throws IOException {
            if (br != null)
                br.close();
        }

        public String next() throws IOException {
            if (i < s.length)
                return s[i++];
            String st = br.readLine();
            while (st == "")
                st = br.readLine();
            s = st.split(regex, 0);
            i = 0;
            return s[i++];
        }

        public int nextInt() throws NumberFormatException, IOException {
            return Integer.parseInt(next());
        }

        public Long nextLong() throws NumberFormatException, IOException {
            return Long.parseLong(next());
        }

        public Double nextDouble() throws NumberFormatException, IOException {
            return Double.parseDouble(next());
        }
    }

    private void solve() {
        try {
            long a;
            a = sc.nextLong();
            long aa = a;

            long[] insu = new long[1000];
            long[] insuCount = new long[1000];
            int idx = 0;
            for (long i = 2; i * i <= aa; i++) {
                while (a % i == 0) {
                    a /= i;
                    insu[idx] = i;
                    insuCount[idx]++;

                }
                if (insu[idx] == i)
                    idx++;
            }
            int count = 0;
            if (a != 1) {
                count++;
            }
            for (int i = 0; i < idx; i++) {
                for (int j = 1; j <= insuCount[i]; j++) {
                    insuCount[i] -= j;
                    count++;
                }
            }
            System.out.println(count);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

}
