IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.

A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.

Output

Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.

Examples

Input

12


Output

2
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
template <class T>
void splitStr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
inline long long two(long long n) { return 1 << n; }
inline long long isOnBit(long long n, long long b) { return (n >> b) & 1; }
inline void onBit(long long &n, long long b) { n |= two(b); }
inline void offBit(long long &n, long long b) { n &= ~two(b); }
inline long long lastBit(long long n) { return n & (-n); }
inline long long cntBit(long long n) {
  long long res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
const long long dx4[] = {-1, 0, 1, 0};
const long long dy4[] = {0, 1, 0, -1};
const long long dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const long long dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modexp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(15);
  mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
  long long n, t;
  t = 1;
  while (t--) {
    cin >> n;
    long long a = n / 2, b = n / 3, c = n / 5, d = n / 7;
    long long e = n / 6, f = n / 14, g = n / 15, h = n / 35;
    long long x = a + b + c + d - e - f - g - h;
    x -= n / 10;
    x -= n / 21;
    x += n / 30 + n / 42 + n / 70 + n / 105;
    x -= n / 210;
    cout << n - x << '\n';
  }
}
