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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;

public class A {
        private static StreamTokenizer in;
        private static PrintWriter out;

        private static int nextInt() throws Exception {
                in.nextToken();
                return (int) in.nval;
        }

        private static String nextString() throws Exception {
                in.nextToken();
                return in.sval;
        }

        public static void main(String[] args) throws Exception {
                in = new StreamTokenizer(new BufferedReader(new InputStreamReader(
                                System.in)));
                out = new PrintWriter(System.out);

                int NN = 3000, N = 55, n = 8;
                boolean[] sieve = new boolean[N + 1];
                Arrays.fill(sieve, true);
                sieve[0] = sieve[1] = false;
                for (int i = 2; i <= n; i++) {
                        if (sieve[i]) {
                                for (int j = i * i; j <= N; j += i) {
                                        sieve[j] = false;
                                }
                        }
                }

                int[] colls = new int[NN + 1];
                for (int i = 1; i <= NN; i++) {
                        int c = 0;
                        int ii = i;
                        for (int j = 2; j <= N; j++) {
                                if (sieve[j]) {
                                        if (ii % j == 0) {
                                                while (ii % j == 0) {
                                                        ii /= j;
                                                }
                                                c++;
                                        }
                                }
                        }

                        if (ii == 1 && c == 2) {
                                colls[i]++;
                        }
                        if (ii > 1 && c == 1) {
                        	colls[i]++;
                        }
                }

                for (int i = 1; i <= NN; i++) {
                        colls[i] += colls[i - 1];
                }

                out.println(colls[nextInt()]);

                out.flush();

        }
}