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 MAXN = 4000000;
long long f[MAXN], g[MAXN];
inline long long solve(long long n) {
  long long m = 1;
  for (; m * m <= n; ++m) {
    f[m] = n / m - 1;
  }
  for (int i = 1; i < (int)m; ++i) {
    g[i] = i - 1;
  }
  for (int i = 2; i < m; ++i) {
    if (g[i] == g[i - 1]) {
      continue;
    }
    for (int j = 1; j < m && j <= n / i / i; ++j) {
      if (1LL * i * j < m) {
        f[j] -= f[i * j] - g[i - 1];
      } else {
        f[j] -= g[n / i / j] - g[i - 1];
      }
    }
    for (int j = m - 1; j >= 1LL * i * i; --j) {
      g[j] -= g[j / i] - g[i - 1];
    }
  }
  long long ans = 0;
  for (int i = 2; i <= n / i / i; ++i) {
    ans += g[i] != g[i - 1];
  }
  for (int i = 2; i <= n / i; ++i) {
    if (g[i] != g[i - 1]) {
      ans += f[i] - g[i];
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  while (cin >> n) {
    cout << solve(n) << endl;
  }
}