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;
long long int n, g[320000], f[320000];
long long int sum(long long int n) {
  long long int m = 1, e;
  for (; m * m <= n; ++m) f[m] = n / m - 1;
  for (int i = 1; i <= m; ++i) g[i] = i - 1;
  for (long long int i = 1; i <= m; ++i)
    if (g[i] != g[i - 1]) {
      e = ((n / (1ll * i * i)) < (m - 1) ? (n / (1ll * i * i)) : (m - 1));
      for (long long int j = 1; j <= e; ++j)
        if (i * j <= m - 1)
          f[j] -= f[i * j] - g[i - 1];
        else
          f[j] -= g[n / (i * j)] - g[i - 1];
      for (long long int j = m; j >= i * i; --j) g[j] -= g[j / i] - g[i - 1];
    }
  long long int ans = 0;
  for (int i = 1; i < m; ++i)
    if (g[i] != g[i - 1]) ans += f[i] - g[i];
  return ans;
}
int main() {
  cin >> n;
  int k = pow(n, 1.0 / 3);
  cout << sum(n) + g[k] << '\n';
  return 0;
}