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
//package CodeForces;

import java.io.*;
import java.util.*;

public class Main implements Runnable {
    boolean[] prime = new boolean[3001];
    private void solve() throws IOException {
        Arrays.fill (prime, true);
        prime[1] = false;
        for (int i = 3; i <= 3000; i++) {            
            for (int j = 2; j <= Math.sqrt(i); j++) {
                if (i % j == 0) {
                    prime[i] = false;
                    break;
                }
            }
        }
        int n = nInt();
        int num = 0;
        for (int i = 1; i <= n; i++) {
            if (alm (i)) {
                num++;
               // out.println (i);
            }
        }
        out.println (num);
    }
    boolean alm(int n) {
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            if ((n % i == 0) && (prime[i])) cnt++;
        }
        return cnt == 2;
    }
    
    //-------------------------------

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

    BufferedReader in;
    StringTokenizer tokenizer;
    PrintWriter out;

    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            out = new PrintWriter(System.out);
            solve();
            in.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nInt() throws IOException {
        return Integer.parseInt(nToken());
    }

    long nLong() throws IOException {
        return Long.parseLong(nToken());
    }

    double nDouble() throws IOException {
        return Double.parseDouble(nToken());
    }

    String nToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(in.readLine());
        }
        return tokenizer.nextToken();
    }
}

class StopWatch {
    private final long start = System.currentTimeMillis();
    public double elapsedTime() {
        long now = System.currentTimeMillis();
        return (now - start) / 1000.;
    }
}