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;
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  if (b == 0)
    return a;
  else
    return gcd(a % b, b);
}
int main() {
  long long int i, j;
  long long int n;
  cin >> n;
  long long int max = 0;
  long long int min = 0;
  if (n % 7 == 0) {
    max = (n / 7) * 2;
    min = (n / 7) * 2;
  } else {
    long long int x = n % 7;
    if (x > 2) {
      max = (n / 7) * 2 + 2;
      if (x > 5) {
        min = (n / 7) * 2 + (7 - x);
      } else {
        min = (n / 7) * 2;
      }
    } else {
      max = (n / 7) * 2 + x;
      min = (n / 7) * 2;
    }
  }
  cout << min << " " << max << endl;
  return 0;
}
