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
/**
 * Created by IntelliJ IDEA.
 * User: user
 * Date: 19.08.2010
 * Time: 22:16:12
 * To change this template use File | Settings | File Templates.
 */

import java.io.*;

public class Simple {
    
    static long pow(long a, int n)
    {
        if(n==0)
            return 1;
        else
        {
            long r=a;
            for(int i=1;i<n;i++)
                r*=a;

            return r;
        }
    }


    public static void main(String[] argv) throws IOException {
        BufferedReader in = new  BufferedReader(new InputStreamReader(System. in));

        int n = Integer.parseInt(in.readLine());

        if(n<6)
        {
            System.out.println(0);
            return;
        }

        int[] t = new int[n];


        int ts=3;
        t[0] = 1;
        t[1] = 2;
        t[2] = 3;

        for(int i=4;i<=n;i++)
        {
            boolean f=false;
            for(int j=2;j<i;j++)
            {
                if( i%j == 0)
                {
                    f=false;
                    break;
                }
                else
                    f=true;

            }

            if(f)
                t[ts++]=i;
        }

        // print result of filling
//        for(int i=0;i<ts;i++)
//            System.out.printf("%d ",t[i]);
//        System.out.println();


        int tz=ts-1;
        int c = 0;
        for(int y=1; y<tz; y++)
        {
            for(int x=y+1; x<tz; x++)
            {
                for(int a=1; a<10; a++)
                    for(int b=1; b < 10; b++)
                    {
                        long k= pow(t[x],a) * pow(t[y],b);
                        if( k <= n && k>0)
                            c++;
                    }
            }
        }
        System.out.println(c);


    }

}
