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 p1():
	n = int(input())
	for i in range(0,n,1234567):
		for j in range(0,n-i,123456):
			if (n-i-j)%1234 == 0:
				return 'YES'
	return 'NO'
def p2():
	n,m = [int(i) for i in input().split()]
	l = [int(i) for i in input().split()]
	for i in range(n):
		if m-i-1 < 0 :
			return l[m-1]
		m = m-i-1
def p3():
	n = int(input())
	b = n//7*2+min(2,n%7)
	a = n//7*2
	if n == n//7*7+6:
		a = a+1
	print(a,b)
p3()
