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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.function.IntPredicate;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author AnandOza
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DDivGame solver = new DDivGame();
        solver.solve(1, in, out);
        out.close();
    }

    static class DDivGame {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long n = in.nextLong();

            long answer = 0;
            for (long p = 2; p * p <= n; p++) {
                int c = 0;
                while (n % p == 0) {
                    c++;
                    n /= p;
                }
                if (c == 0)
                    continue;
                int finalC = c;
                answer += -1 + BinarySearch.binarySearch(x -> x * (x + 1) / 2 > finalC);
            }

            if (n > 1) {
                answer++;
            }

            out.println(answer);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

    }

    static class BinarySearch {
        public static int binarySearch(IntPredicate bigEnough) {
            int tooSmall = 0, tooBig;
            if (bigEnough.test(tooSmall)) {
                return tooSmall;
            }

            for (tooBig = 1; !bigEnough.test(tooBig); tooBig *= 2)
                ;

            if (!bigEnough.test(tooBig - 1)) {
                return tooBig;
            }

            while (tooBig - tooSmall > 1) {
                int center = tooSmall + (tooBig - tooSmall) / 2;
                if (!bigEnough.test(center)) {
                    tooSmall = center;
                } else {
                    tooBig = center;
                }
            }

            return tooBig;
        }

    }
}

