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.
def solve(n):
  if n == 1:
    return 0,1
  if n <= 5:
    return 0,2
  if n< 7:
    return 1,2
  if n == 7:
    return 2,2
  rem = n % 7
  if not rem :
    return (n//7)*2,(n//7)*2
  else:
    b = (n // 7)*2
    if rem == 1:
      return b, b+1
    if rem <= 5:
      return b,b+2
    if rem< 7:
      return b+1,b+2

      
  

def main() :
  n = int(input())
  print(*solve(n))
main()
