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():
	# take in input

	n = int(raw_input())
	maximum = 2*(int(n/7))
	if n%7 == 1:
		maximum += 1
	elif n%7 == 0:
		pass
	else: 
		maximum += 2
		
	minimum = 2*(int(n/7))
	if n%7 == 6:
		minimum += 1

	print "%d %d" % (minimum, maximum)
		



if __name__ == "__main__":
	main()