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 count_weekends(n):
    min_weekends = 0
    max_weekends = 0
    for i in range(1, n+1):
        if i%7 == 6 or i%7 == 0:
            min_weekends += 1
        if i%7 == 1 or i%7 == 2:
            max_weekends += 1
    return "%d %d" % (min_weekends, max_weekends)

n = int(input())
# n = 10**6
# from time import time
# start = time()
print(count_weekends(n))
# print(time() - start)
