#include <fstream>
using namespace std;

ifstream cin ("ciur.in");
ofstream cout("ciur.out");

int prim[1000001 / 31 + 1];

int main()
{
    int n, l = 0;
    cin >> n;
    for(int i = 2; i <= n; ++i) prim[i >> 5] |= (1 << (i & 31));
    for(int i = 4; i <= n; i += 2) prim[i >> 5] &=~ (1 << (i & 31));
    for(int i = 3; i * i <= n; i += 2)
        if(prim[i >> 5] & (1 << (i & 31)))
            for(int j = i * i; j <= n; j += (i << 1)) prim[j >> 5] &=~ (1 << (j & 31));
    for(int i = 2; i <= n; ++i)
        if(prim[i >> 5] & (1 << (i & 31))) l++;
    cout << l;
    return 0;
}