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 N = 5e6 + 2;
bool np[N];
long long prime[N], pi[N];
int cnt = 0;
void getprime() {
  np[0] = np[1] = true;
  pi[0] = pi[1] = 0;
  for (int i = 2; i < N; ++i) {
    if (!np[i]) prime[++cnt] = i;
    pi[i] = cnt;
    for (int j = 1; j <= cnt && i * prime[j] < N; ++j) {
      np[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int phi[PM + 1][M + 1], sz[M + 1];
void init() {
  getprime();
  sz[0] = 1;
  for (int i = 0; i <= PM; ++i) phi[i][0] = i;
  for (int i = 1; i <= M; ++i) {
    sz[i] = prime[i] * sz[i - 1];
    for (int j = 1; j <= PM; ++j)
      phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
  }
}
int sqrt2(long long x) {
  long long r = (long long)sqrt(x - 0.1);
  while (r * r <= x) ++r;
  return int(r - 1);
}
int sqrt3(long long x) {
  long long r = (long long)cbrt(x - 0.1);
  while (r * r * r <= x) ++r;
  return int(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 <= prime[s] * prime[s]) return pi[x] - s + 1;
  if (x <= prime[s] * prime[s] * prime[s] && x < N) {
    int s2x = pi[sqrt2(x)];
    long long ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
    for (int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];
    return ans;
  }
  return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
}
long long getpi(long long x) {
  if (x < N) return pi[x];
  long long ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
  for (int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i)
    ans -= getpi(x / prime[i]) - i + 1;
  return ans;
}
long long lehmer_pi(long long x) {
  if (x < N) return pi[x];
  int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
  int b = (int)lehmer_pi(sqrt2(x));
  int c = (int)lehmer_pi(sqrt3(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 / prime[i];
    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 / prime[j]) - (j - 1);
  }
  return sum;
}
int main() {
  init();
  long long n;
  scanf("%lld", &n);
  long long tp = cbrt(n * 1.0);
  long long sum = lehmer_pi(tp);
  for (int i = 1; i <= cnt; i++) {
    if (prime[i] * prime[i] > n) break;
    long long tp = n / prime[i];
    long long s = lehmer_pi(tp);
    s -= lehmer_pi(prime[i]);
    sum += s;
  }
  cout << sum << endl;
  return 0;
}