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.
from sys import stdin,stdout
r,w = stdin.readline,stdout.write

d = {1:(0,1),2:(0,2),3:(0,2),4:(0,2)}
n = int(r(),10)

if n in d:
	w("%i %i" %(d[n][0],d[n][1]))
	exit()

if not n%7:
	w('%i %i' %((n/7)*2,(n/7)*2))
	exit()
	
mult_prox = 7*round(n/7.)

t = (mult_prox/7)*2

if n == mult_prox+1:
	w('%i %i' %(t,t+1))
elif n == mult_prox+2:
	w('%i %i' %(t,t+2))
elif n == mult_prox-1:
	w('%i %i' %(t-1,t))
elif n == mult_prox-2:
	w('%i %i' %(t-2,t))
else:
	mult_prox = 7*n/7
	t = ((mult_prox+7)/7)*2
	w('%i %i' %(t-2,t))