On the planet Mars a year lasts exactly n days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 1 000 000) — the number of days in a year on Mars.

Output

Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.

Examples

Input

14


Output

4 4


Input

2


Output

0 2

Note

In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .

In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
std::ostream &operator<<(std::ostream &out, const std::vector<T> &v) {
  if (!v.empty()) {
    out << '[';
    std::copy(v.begin(), v.end(), std::ostream_iterator<T>(out, ", "));
    out << "\b\b]";
  }
  return out;
}
template <typename T1, typename T2>
std::ostream &operator<<(std::ostream &out, const std::pair<T1, T2> &p) {
  out << "[" << p.first << ", " << p.second << "]";
  return out;
}
template <class T, class U>
void chmin(T &t, U f) {
  if (t > f) t = f;
}
template <class T, class U>
void chmax(T &t, U f) {
  if (t < f) t = f;
}
template <typename T>
void uniq(vector<T> &v) {
  v.erase(unique(v.begin(), v.end()), v.end());
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  int rest = N % 7;
  int must = N / 7;
  int mi = must * 2;
  if (rest == 6) {
    mi++;
  }
  int ma = must * 2;
  if (rest == 0) {
  } else if (rest == 1) {
    ma++;
  } else {
    ma += 2;
  }
  cout << mi << " " << ma << endl;
}
