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 f[340000], g[340000], n, k;
long long dp(long long n) {
  long long i, j, m, res = 0;
  for (m = 1; m * m <= n; ++m) f[m] = n / m - 1;
  for (i = 1; i <= m; ++i) g[i] = i - 1;
  for (i = 2; i <= m; ++i) {
    if (g[i] == g[i - 1]) continue;
    for (j = 1; j <= min(m - 1, n / i / i); ++j) {
      if (i * j < m)
        f[j] -= f[i * j] - g[i - 1];
      else
        f[j] -= g[n / i / j] - g[i - 1];
    }
    for (j = m; j >= i * i; --j) g[j] -= g[j / i] - g[i - 1];
  }
  for (i = 2; i < m; ++i) {
    if (g[i] == g[i - 1]) continue;
    res += f[i] - g[i];
  }
  return res;
}
int main() {
  scanf("%I64d", &n);
  k = pow(n, 1.0 / 3) + 1e-9;
  printf("%I64d\n", dp(n) + g[k]);
}