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 = 320000;
long long n, m;
long long f[maxn], g[maxn];
long long calc(long long x) {
  while (m * m <= n) ++m;
  for (int i = 1; i <= m; i++) f[i] = n / i - 1;
  for (int i = 1; i <= m; i++) g[i] = i - 1;
  for (int i = 2; i <= m; i++)
    if (g[i] ^ g[i - 1]) {
      for (int j = 1; j <= min(n / i / i, m * 1ll); j++)
        if (i * 1ll * j <= m)
          f[j] -= f[i * j] - g[i - 1];
        else
          f[j] -= g[n / i / j] - g[i - 1];
      for (int j = m; j >= i * 1ll * i; j--) g[j] -= g[j / i] - g[i - 1];
    }
  long long ans = 0;
  for (int i = 2; i < m; i++) ans += (g[i] != g[i - 1]) * (f[i] - g[i]);
  return ans;
}
int main() {
  scanf("%I64d", &n);
  int t = cbrt(n);
  printf("%I64d\n", calc(n) + g[t]);
}