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;
template <class T>
inline void check_min(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void check_max(T& a, T b) {
  if (a < b) a = b;
}
const int MAX_N = 1e6 + 5;
vector<int> primes;
void gen_prime() {
  primes.reserve(MAX_N / 10);
  static bool np[MAX_N];
  for (int i = 2; i < MAX_N; ++i) {
    if (!np[i]) primes.push_back(i);
    for (int p : primes) {
      if (i * p >= MAX_N) break;
      np[i * p] = true;
      if (i % p == 0) break;
    }
  }
}
long long a[MAX_N], b[MAX_N];
int main() {
  gen_prime();
  long long n, x = 0;
  scanf("%lld", &n);
  do ++x, a[x] = x, b[x] = n / x;
  while (x * x <= n);
  for (int p : primes) {
    long long pp = (long long)p * p;
    if (pp > n) break;
    for (long long j = 1; j <= x && n / j >= pp; ++j) {
      long long o = j * p;
      b[j] -= (o >= x ? a[n / o] : b[o]) - a[p - 1];
    }
    for (long long j = x; j >= pp; --j) {
      a[j] -= a[j / p] - a[p - 1];
    }
  }
  long long ans = 0;
  for (int p : primes) {
    ans += (long long)p * p * p <= n;
    if (p < n / p)
      ans += b[p] - a[p];
    else
      break;
  }
  printf("%lld\n", ans);
  return 0;
}