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 <class X>
void input(vector<X>& a, int N) {
  X temp;
  for (long long int i = 0; i < N; i++) {
    cin >> temp;
    a.push_back(temp);
  }
}
void solve() {
  int n;
  cin >> n;
  int ans = (n / 7) * 2;
  n %= 7;
  if (n == 6) {
    cout << ans + 1 << " " << ans + ((2) < (n) ? (2) : (n));
    return;
  }
  cout << ans << " " << ans + ((2) < (n) ? (2) : (n));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
