Problem: If an integer a is divisible by another integer b, then b is called the divisor of a.

For example: 12 has positive 6 divisors. They are 1, 2, 3, 4, 6 and 12.

Let’s define a function D(n) — number of integers between 1 and n (inclusive) which has exactly four positive divisors.

Between 1 and 10 only the integers 6, 8 and 10 has exactly four positive divisors. So, D(10) = 3.

You are given an integer n. You have to calculate D(n).

Input

The only line contains integer n (1 ≤ n ≤ 1011) — the parameter from the problem statement.

Output

Print the only integer c — the number of integers between 1 and n with exactly four divisors.

Examples

Input

10


Output

3


Input

20


Output

5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e6 + 5;
const int M = 7;
vector<int> lp, primes, pi;
int phi[MAX + 1][M + 1], sz[M + 1];
void factor_sieve() {
  lp.resize(MAX);
  pi.resize(MAX);
  lp[1] = 1;
  pi[0] = pi[1] = 0;
  for (int i = 2; i < MAX; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      primes.emplace_back(i);
    }
    for (int j = 0; j < primes.size() && primes[j] <= lp[i]; j++) {
      int x = i * primes[j];
      if (x >= MAX) break;
      lp[x] = primes[j];
    }
    pi[i] = primes.size();
  }
}
void init() {
  factor_sieve();
  for (int i = 0; i <= MAX; i++) {
    phi[i][0] = i;
  }
  sz[0] = 1;
  for (int i = 1; i <= M; i++) {
    sz[i] = primes[i - 1] * sz[i - 1];
    for (int j = 1; j <= MAX; j++) {
      phi[j][i] = phi[j][i - 1] - phi[j / primes[i - 1]][i - 1];
    }
  }
}
int sqrt2(long long x) {
  long long r = sqrt(x - 0.1);
  while (r * r <= x) ++r;
  return r - 1;
}
int cbrt3(long long x) {
  long long r = cbrt(x - 0.1);
  while (r * r * r <= x) ++r;
  return r - 1;
}
long long getphi(long long x, int s) {
  if (s == 0) return x;
  if (s <= M) {
    return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
  }
  if (x <= primes[s - 1] * primes[s - 1]) {
    return pi[x] - s + 1;
  }
  if (x <= primes[s - 1] * primes[s - 1] * primes[s - 1] && x < MAX) {
    int sx = pi[sqrt2(x)];
    long long ans = pi[x] - (sx + s - 2) * (sx - s + 1) / 2;
    for (int i = s + 1; i <= sx; ++i) {
      ans += pi[x / primes[i - 1]];
    }
    return ans;
  }
  return getphi(x, s - 1) - getphi(x / primes[s - 1], s - 1);
}
long long getpi(long long x) {
  if (x < MAX) return pi[x];
  int cx = cbrt3(x), sx = sqrt2(x);
  long long ans = getphi(x, pi[cx]) + pi[cx] - 1;
  for (int i = pi[cx] + 1, ed = pi[sx]; i <= ed; i++) {
    ans -= getpi(x / primes[i - 1 - 1]) - i + 1;
  }
  return ans;
}
long long lehmer_pi(long long x) {
  if (x < MAX) return pi[x];
  int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
  int b = (int)lehmer_pi(sqrt2(x));
  int c = (int)lehmer_pi(cbrt3(x));
  long long sum = getphi(x, a) + (long long)(b + a - 2) * (b - a + 1) / 2;
  for (int i = a + 1; i <= b; i++) {
    long long w = x / primes[i - 1];
    sum -= lehmer_pi(w);
    if (i > c) continue;
    long long lim = lehmer_pi(sqrt2(w));
    for (int j = i; j <= lim; j++) {
      sum -= lehmer_pi(w / primes[j - 1]) - (j - 1);
    }
  }
  return sum;
}
long long power(long long a, long long b) {
  long long x = 1, y = a;
  while (b) {
    if (b & 1) x = x * y;
    y = y * y;
    b >>= 1;
  }
  return x;
}
void solve(long long n) {
  long long ans = 0;
  for (int i = 0; i < (int)primes.size(); i++) {
    if (power(primes[i], 3) <= n) {
      ans += 1;
    } else {
      break;
    }
  }
  long long cnt = 0;
  long long q = 0;
  for (int i = 0; i < (int)primes.size(); i++) {
    long long x = (long long)primes[i];
    q = n / x;
    if (q <= x) continue;
    if (q == 0) continue;
    cnt = lehmer_pi(q);
    if (q >= primes[i]) {
      cnt -= lehmer_pi(x);
    }
    if (cnt <= 0) break;
    ans += cnt;
  }
  std::cout << ans << '\n';
}
int main(int argc, char const *argv[]) {
  long long n;
  init();
  scanf("%lld", &n);
  solve(n);
  return 0;
}