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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using uint8 = unsigned char;
using uint16 = unsigned short int;
using uint32 = unsigned int;
using uint64 = unsigned long long;
using int8 = signed char;
using int16 = short int;
using int32 = int;
using int64 = long long;
using pi32 = std::pair<int32, int32>;
using pi64 = std::pair<int64, int64>;
namespace oct {
const int tn4[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const int tn8[8][2] = {{1, 0},  {1, 1},   {0, 1},  {-1, 1},
                       {-1, 0}, {-1, -1}, {0, -1}, {1, -1}};
const double exp = 1e-8;
void sync(int pre = 8);
template <typename _Tp>
_Tp &mad(_Tp &x, _Tp y);
template <typename _Tp>
_Tp &mid(_Tp &x, _Tp y);
template <typename _Tp>
bool in(_Tp x, _Tp l, _Tp r);
template <typename _Tp>
_Tp sqr(_Tp x);
template <typename _Tp>
_Tp power(_Tp x, int64 m);
template <typename _Tp>
void sort(_Tp &x, _Tp &y);
template <typename _Tp1, typename _Tp2>
std::pair<_Tp1, _Tp2> operator+(std::pair<_Tp1, _Tp2> x,
                                std::pair<_Tp1, _Tp2> y);
template <typename _Tp>
_Tp &operator+=(_Tp &x, _Tp y);
inline void sync(int pre) {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cout.flags(std::ios::fixed);
  std::cout.precision(pre);
  std::cout.setf(std::ios::showpoint);
}
template <typename _Tp>
inline _Tp &mad(_Tp &x, _Tp y) {
  return x = std::max(x, y);
}
template <typename _Tp>
inline _Tp &mid(_Tp &x, _Tp y) {
  return x = std::min(x, y);
}
template <typename _Tp>
inline bool in(_Tp x, _Tp l, _Tp r) {
  return l <= x && x <= r;
}
template <typename _Tp>
inline _Tp sqr(_Tp x) {
  return x * x;
}
template <typename _Tp>
inline _Tp power(_Tp x, int64 m) {
  return m == 1 ? x : (m & 1 ? power(x * x, m / 2) * x : power(x * x, m / 2));
}
template <typename _Tp>
inline void sort(_Tp &x, _Tp &y) {
  if (x > y) std::swap(x, y);
}
template <typename _Tp1, typename _Tp2>
inline std::pair<_Tp1, _Tp2> operator+(std::pair<_Tp1, _Tp2> x,
                                       std::pair<_Tp1, _Tp2> y) {
  return x.first += y.first, x.second += y.second, x;
}
template <typename _Tp>
_Tp &operator+=(_Tp &x, _Tp y) {
  return x = x + y;
}
}  // namespace oct
const int N = 316228;
const int M = 2 * N;
const int P = 27294;
void euler_prime();
bool vis[N];
int lim, cnt;
int p[P];
int sum[N];
int last[M];
int64 n;
int64 val[M], f[M];
inline void euler_prime() {
  for (register int i = 2; i <= lim; ++i) {
    if (!vis[i]) p[++p[0]] = i;
    sum[i] = sum[i - 1] + !vis[i];
    for (register int j = 1; j <= p[0] && i * p[j] <= lim; ++j) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int main() {
  oct::sync();
  std::cin >> n;
  lim = sqrt(n);
  euler_prime();
  for (register int64 i = 1; i <= n; i = n / (n / i) + 1) val[++cnt] = n / i;
  std::reverse(val + 1, val + cnt + 1);
  std::copy(val + 1, val + cnt + 1, f + 1);
  for (register int i = 1; i <= p[0]; ++i) {
    for (register int j = cnt; j; --j) {
      int64 k = val[j] / p[i], pos = k <= lim ? k : cnt + 1 - n / k;
      if (k < p[i]) break;
      f[j] -= f[pos] + last[pos] - i + 1;
      last[j] = i;
    }
  }
  int64 ans = 0;
  for (register int i = 1; i <= cnt; ++i) f[i] += last[i] - 1;
  for (register int i = 1; i <= p[0]; ++i) ans += f[cnt + 1 - p[i]] - i;
  if (n != 1) {
    ans += std::upper_bound(p + 1, p + p[0] + 1, floor(pow(n, 1. / 3))) - p - 1;
  }
  std::cout << ans << std::endl;
  return 0;
}