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 main():
	n = int(input())
	(minimum, maximum) = solver(n)
	print(minimum, maximum)

def solver(n):
	weeks = n // 7
	daysOff = weeks * 2
	daysLeft = n - weeks * 7	
	if daysLeft <= 1:
		maximum = daysOff + daysLeft
	else:
		maximum = daysOff + 2
	if daysLeft >= 6:
		minimum = daysOff + daysLeft - 5
	else:
		minimum = daysOff
	return (minimum, maximum)

#print(solver(2))
main()