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>
int main(void) {
  long n;
  long mn, mx;
  scanf("%ld", &n);
  if ((n / 7) >= 1) {
    if (n % 7 == 0) {
      mn = 2 * (n / 7);
      mx = mn;
    } else if (n % 7 == 1) {
      mn = 2 * (n / 7);
      mx = mn + 1;
    } else if (n % 7 >= 2 && n % 7 <= 5) {
      mn = 2 * (n / 7);
      mx = mn + 2;
    } else if (n % 7 == 6) {
      mn = 2 * (n / 7) + 1;
      mx = mn + 1;
    }
  } else {
    if (n % 7 == 0 || n % 7 == 1) {
      mn = 0;
      mx = mn + n;
    } else if (n % 7 >= 2 && n % 7 <= 5) {
      mn = 0;
      mx = mn + 2;
    } else if (n % 7 == 6) {
      mn = 1;
      mx = mn + 1;
    }
  }
  printf("%ld %ld", mn, mx);
  return 0;
}
