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 high[340000], low[340000];
long long fun(long long n) {
  long long i, m, ret = 0, p, s, x;
  for (m = 1; m * m <= n; m++) high[m] = n / m - 1;
  for (i = 1; i <= m; i++) low[i] = i - 1;
  for (p = 2; p <= m; p++) {
    if (low[p] == low[p - 1]) continue;
    s = min(n / (p * p), m - 1);
    for (x = 1; x <= s; x++) {
      if (x * p <= m - 1)
        high[x] -= high[x * p] - low[p - 1];
      else
        high[x] -= low[n / (x * p)] - low[p - 1];
    }
    for (x = m; x >= p * p; x--) low[x] -= low[x / p] - low[p - 1];
  }
  for (p = 2; p < m; p++) {
    if (low[p] == low[p - 1]) continue;
    ret += high[p] - low[p];
  }
  return ret;
}
int main() {
  long long n, k;
  cin >> n;
  k = pow(n, 1.0 / 3) + 1e-9;
  cout << fun(n) + low[k] << endl;
}