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;
void fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool find(int x, vector<int> v) {
  for (int i = 0; i < v.size(); i++) {
    if (x == v[i]) return 1;
  }
  return 0;
}
bool finds(string x, char y) {
  for (int i = 0; i < x.size(); i++) {
    if (x[i] == y) return 1;
  }
  return 0;
}
int tonum(string x) {
  stringstream s;
  s << x;
  int a;
  s >> a;
  return a;
}
int main() {
  fast();
  int n;
  cin >> n;
  int x = (n / 7) * 2;
  if (n % 7 == 6) x++;
  int y = (n / 7) * 2;
  if (n % 7 >= 2)
    y += 2;
  else if (n % 7 == 1)
    y++;
  cout << x << " " << y;
  return 0;
}
