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.
n=int(input())
dmin=0
dmax=0

def min(n):
    if n%7==6:
        dmin=(n//7)*2+1
        return dmin
    elif n%7==0:
        return (n//7)*2
    else:
        dmin=(n//7)*2
        return dmin
x=min(n)
def max(n,x):
    if n%7==1:
        dmax=x+1
        return dmax
    elif n%7>=2:
        if n%7==6:
            dmax=x+1
            return dmax
        else:
            dmax=x+2
            return dmax
    elif n%7==0:
        return (n//7)*2
y=max(n,x)
print(str(x)+" "+str(y))