A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and n, inclusive.

Input

Input contains one integer number n (1 ≤ n ≤ 3000).

Output

Output the amount of almost prime numbers between 1 and n, inclusive.

Examples

Input

10


Output

2


Input

21


Output

8
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.List;
import java.util.Queue;
import java.awt.*;

public class codefors implements Runnable {

    private BufferedReader br = null;
    private PrintWriter pw = null;
    private StringTokenizer stk = new StringTokenizer("");

    public static void main(String[] args) {
        new Thread(new codefors()).run();
    }

    public void run() {
        /*
         * try { // br = new BufferedReader(new FileReader("input.txt")); pw =
         * new PrintWriter("output.txt"); } catch (FileNotFoundException e) {
         * e.printStackTrace(); }
         */
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        solver();
        pw.close();

    }

    private void nline() {
        try {
            if (!stk.hasMoreTokens())
                stk = new StringTokenizer(br.readLine());
        } catch (IOException e) {
            throw new RuntimeException("KaVaBUnGO!!!", e);
        }
    }

    private String nstr() {
        while (!stk.hasMoreTokens())
            nline();
        return stk.nextToken();
    }

    private int ni() {
        return Integer.valueOf(nstr());

    }

    private double nd() {
        return Double.valueOf(nstr());

    }

    private BigInteger nbi() {
        return new BigInteger(nstr());
    }

    String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
        }
        return null;
    }

    boolean prime_check(int k) {
        for (int i = 2; i < Math.sqrt(k) + 1; i++) {
            if (k % i == 0)
                return false;
        }
        return true;
    }

    private void solver() {
        int n = ni();
        ArrayList<Integer> ar = new ArrayList<Integer>();
        ar.add(2);
        for (int i = 2; i < n; i++) {
            if (prime_check(i))
                ar.add(i);
        }

        ArrayList<Integer> ar2 = new ArrayList<Integer>();
        for (int i = 0; i < ar.size() - 1; i++) {
            for (int j = i + 1; j < ar.size(); j++) {
                if (ar.get(i) * ar.get(j) <= n)
                    ar2.add(ar.get(i) * ar.get(j));

            }
        }

        int ans = 0;
        for (int i = 1; i <= n; i++) {
            int k = 0;
            for (int j = 0; j < ar2.size(); j++) {
                if (i % ar2.get(j) == 0) {
                    k++;
                }
            }
            if (k == 1) {
                ans++;
            }
        }

        System.out.println(ans);
    }

    void exit() {
        System.exit(0);
    }
}
