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

def getOff(n, i):
    t = 0
    while i <= 7 and n > 0:
        if i == 6 or i == 7:
            t += 1
        i += 1
        n -= 1
    t += (n // 7) * 2
    n = n % 7
    if n == 6:
        t += 1
    return t

def main():
    n = int(stdin.readline())
    maxOff, minOff = 0, n
    for i in range(1, 8):
        t = getOff(n, i)
        maxOff = max(maxOff, t)
        minOff = min(minOff, t)
    print(minOff, maxOff)

main()
