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
#	✪ H4WK3yE乡
#	Mayank Chaudhary
#	ABES EC , Ghaziabad
# ///==========Libraries, Constants and Functions=============///
import sys
from bisect import bisect_left,bisect_right
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log
from itertools import permutations
inf = float("inf")
mod = 1000000007
mini=1000000007

def fact(n):                                                             # <------  To calculate factorial of n under modulo m
    if n==0:
        return 1
    p=1;d=(10**9)+7
    for i in range(1,n+1):
        p=p*i
        p=p%d
    return p
def ncr(n,r):                                                          # < ------ To calculate nCr mod p value using Fermat Little under modulo m
    d=10**9+7
    num=fact(n)
    den=(fact(r)*fact(n-r))%d
    den=pow(den,d-2,d)
    return (num*den)%d
def sieve(n):                                                          # <----- sieve of eratosthenes for prime no.
    prime=[True for i in range(n+1)]
    lst=[]
    p=2
    while p*p<=n:
        if prime[p]:
            for i in range(p*p,n+1,p):
                prime[i]=False
        p=p+1
    for i in range(2,n+1):
        if prime[i]:
            lst.append(i)
    return lst
        
def binary(number):                                             #  <----- calculate the no. of 1's in binary representation of number
    result=0
    while number:
        result=result+1
        number=number&(number-1)
    return result

def calculate_factors(n):                                #<---- most efficient method to calculate no. of factors of number             
    hh = [1] * (n + 1); 
    p = 2; 
    while((p * p) < n): 
        if (hh[p] == 1): 
            for i in range((p * 2), n, p): 
                hh[i] = 0; 
        p += 1;  
    total = 1; 
    for p in range(2, n + 1): 
        if (hh[p] == 1):  
            count = 0; 
            if (n % p == 0): 
                while (n % p == 0): 
                    n = int(n / p); 
                    count += 1; 
                total *= (count + 1); 
    return total;

def prime_factors(n):
    i = 2
    factors = set()
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.add(i)
    if n > 1:
        factors.add(n)
    if len(factors)==2:
        return True
    else:
        return False

def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///==========MAIN=============///
n=int(input())
count=0
for i in range(1,n+1):
    if prime_factors(i)==True:
        count=count+1
print(count)
