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;
struct s {
  long long mn, mx;
};
s arr[1000009];
int main() {
  long long n, i;
  arr[1].mn = 0;
  arr[1].mx = 1;
  arr[2].mn = 0;
  arr[2].mx = 2;
  arr[3].mn = 0;
  arr[3].mx = 2;
  arr[4].mn = 0;
  arr[4].mx = 2;
  arr[5].mn = 0;
  arr[5].mx = 2;
  arr[6].mn = 1;
  arr[6].mx = 2;
  arr[7].mn = 2;
  arr[7].mx = 2;
  for (i = 8; i <= 1000000; i++) {
    arr[i].mn = arr[i - 7].mn + 2;
    arr[i].mx = arr[i - 7].mx + 2;
  }
  while (cin >> n) {
    cout << arr[n].mn << " " << arr[n].mx << endl;
  }
}
