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 solve(long n) {
  long div, rem;
  div = n / 7;
  rem = n % 7;
  long max, min;
  int surplus{0};
  if (rem >= 2) {
    surplus = 2;
  } else if (rem >= 1) {
    surplus = 1;
  } else {
    surplus = 0;
  }
  max = div * 2 + surplus;
  surplus = 0;
  if (rem >= 6) {
    surplus = 1;
  } else {
    surplus = 0;
  }
  min = div * 2 + surplus;
  cout << min << " " << max << "\n";
}
int main() {
  long n;
  cin >> n;
  solve(n);
  return 0;
}
