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.
#!/usr/bin/env python
#-*- coding:utf-8 -*-

import sys, math, random, operator
from string import ascii_lowercase
from string import ascii_uppercase
from fractions import Fraction, gcd
from decimal import Decimal, getcontext
from itertools import product, permutations, combinations
from Queue import Queue, PriorityQueue
from collections import deque, defaultdict, Counter
getcontext().prec = 100

MOD = 10**9 + 7
INF = float("+inf")

if sys.subversion[0] != "CPython":  # PyPy?
    raw_input = lambda: sys.stdin.readline().rstrip()
pr = lambda *args: sys.stdout.write(" ".join(str(x) for x in args) + "\n")
epr = lambda *args: sys.stderr.write(" ".join(str(x) for x in args) + "\n")
die = lambda *args: pr(*args) ^ exit(0)

read_str = raw_input
read_strs = lambda: raw_input().split()
read_int = lambda: int(raw_input())
read_ints = lambda: map(int, raw_input().split())
read_float = lambda: float(raw_input())
read_floats = lambda: map(float, raw_input().split())

"---------------------------------------------------------------"

n = read_int()

max = 2*(n / 7) + min(2, n % 7)
min = 2*(n / 7)
if n % 7 == 6:
    min += 1
print min, max
