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.*;

public class AlmostPrime {

  static public void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    int N = Integer.parseInt(in.readLine());
    int[] primos = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
     31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
     73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
    127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
    179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
    233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
    283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
    353, 359, 367, 373, 379, 383, 389, 397, 401, 409,
    419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
    467, 479, 487, 491, 499, 503, 509, 521, 523, 541,
    547, 557, 563, 569, 571, 577, 587, 593, 599, 601,
    607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
    661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
    739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
    811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
    877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
    947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013,
    1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069,
    1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151,
    1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223,
    1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291,
    1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373,
    1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451,
    1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499};
    int n_casi_primos = 0, bla;
    for(int i = 6; i <= N; i++) {
      bla = 0;
      for(int j = 0; j < primos.length; j++) {
        if(primos[j] > N / 2 || bla > 2) {
          break;
        }
        if(i % primos[j] == 0) {
          bla++;
        }
      }
      if(bla == 2) {
        n_casi_primos++;
      }
    }
    System.out.println(n_casi_primos);
  }
}