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", "unroll-loops", "no-stack-protector")
using namespace std;
using ll = long long;
using ld = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using uint = unsigned int;
using ull = unsigned long long;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
default_random_engine generator;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(8) << fixed;
  cerr << setprecision(8) << fixed;
  int tests = 1;
  for (int test = 1; test <= tests; ++test) {
    solve();
  }
}
class BIT {
 public:
  BIT(int n) { arr.resize(n); }
  void add(int pos, int val) {
    for (int i = pos; i < (int)(arr).size(); i |= i + 1) {
      arr[i] += val;
    }
  }
  int sum(int pos) {
    int res = 0;
    for (int i = pos; i >= 0; i = (i & (i + 1)) - 1) {
      res += arr[i];
    }
    return res;
  }

 private:
  vector<int> arr;
};
namespace count_primes_for_all {
const ll MAXN = 1e11;
const ll K = 5000000;
const ll MAXP = 500000;
const ll MAXQ = 20000000;
int cntPrimes = 0;
int primes[MAXP];
int minPrime[K + 1];
int cnt[K + 1];
void sieve() {
  memset(minPrime, -1, sizeof(minPrime));
  for (int i = 2; i <= K; ++i) {
    if (minPrime[i] == -1) {
      minPrime[i] = cntPrimes;
      primes[cntPrimes++] = i;
    }
    for (int j = 0; j < cntPrimes; ++j) {
      int p = primes[j];
      if (j > minPrime[i] || (ll)i * p > K) {
        break;
      }
      minPrime[p * i] = j;
    }
    cnt[i] = cnt[i - 1] + (minPrime[i] == cntPrimes - 1);
  }
}
struct Query {
  int id, pos;
};
int cntQueries = 0;
int sizes[(MAXN - 1) / K + 2];
vector<ll> dp[(MAXN - 1) / K + 2];
vector<Query> queries[MAXP];
vector<int> byPrime[MAXP];
int ans[MAXQ];
void clear() {
  cntQueries = 0;
  for (int i = 0; i < cntPrimes; ++i) {
    byPrime[i].clear();
    queries[i].clear();
  }
}
ll countPrimes(ll n) {
  clear();
  int layers = (n - 1) / K + 1;
  sizes[layers + 1] = 0;
  for (int i = 1; i <= layers; ++i) {
    int j = 0;
    while (ll(primes[j]) * primes[j] <= n / i) {
      ++j;
    }
    sizes[i] = j + 1;
    dp[i].resize(sizes[i]);
  }
  for (int l = layers; l >= 1; --l) {
    for (int k = 1; k < sizes[l]; ++k) {
      ll nextL = ll(l) * primes[k - 1];
      if (n / nextL <= K) {
        queries[k - 1].push_back({cntQueries++, int(n / nextL)});
      }
    }
  }
  for (int i = 2; i <= K; ++i) {
    byPrime[minPrime[i]].push_back(i);
  }
  BIT bit(K + 1);
  bit.add(1, 1);
  for (int i = cntPrimes - 1; i >= 0; --i) {
    for (int pos : byPrime[i]) {
      bit.add(pos, 1);
    }
    for (auto q : queries[i]) {
      ans[q.id] = bit.sum(q.pos);
    }
  }
  cntQueries = 0;
  for (int l = layers; l >= 1; --l) {
    dp[l][0] = n / l;
    for (int k = 1; k < sizes[l]; ++k) {
      dp[l][k] = dp[l][k - 1];
      ll nextL = ll(l) * primes[k - 1];
      if (n / nextL <= K) {
        dp[l][k] -= ans[cntQueries++];
      } else {
        ll val = dp[nextL][min(k, sizes[nextL]) - 1] - max(0, k - sizes[nextL]);
        dp[l][k] -= val;
      }
    }
  }
  return sizes[1] - 2 + dp[1].back();
}
ll countPrimes(ll n, ll l) {
  if (n / l <= K) {
    return cnt[n / l];
  }
  return sizes[l] - 2 + dp[l].back();
}
}  // namespace count_primes_for_all
namespace fast_sieve {
vector<int> primes;
vector<int> minPrime;
void sieve(int n) {
  minPrime.resize(n + 1);
  for (int i = 2; i <= n; ++i) {
    minPrime[i] = i;
  }
  for (int i = 2; i <= n; ++i) {
    if (minPrime[i] == i) {
      primes.push_back(i);
    }
    for (int p : primes) {
      if (p > minPrime[i] || (ll)i * p > n) {
        break;
      }
      minPrime[p * i] = p;
    }
  }
}
}  // namespace fast_sieve
void solve() {
  ll n;
  cin >> n;
  fast_sieve::sieve(sqrtl(n));
  ll res = 0;
  for (ll p : fast_sieve::primes) {
    if (p * p * p > n) {
      break;
    }
    ++res;
  }
  count_primes_for_all::sieve();
  count_primes_for_all::countPrimes(n);
  for (int i = 0; i < (int)(fast_sieve::primes).size(); ++i) {
    res += count_primes_for_all::countPrimes(n, fast_sieve::primes[i]) - i - 1;
  }
  cout << res << endl;
}