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 python3
import sys
import math
import string

class A:
    def __init__(self, days):
        self.days = int(days)

    def get_days(self):
        return self.days

    def get_binary(self):
        return int(math.ceil(self.days / 7.0)+1) * "0 0 0 0 0 1 1 "
    
    def get_binary_array(self, binary):
        return [
            binary.strip().split(' ')[0:self.days],
            binary.strip().split(' ')[1:self.days + 1],
            binary.strip().split(' ')[2:self.days + 2],
            binary.strip().split(' ')[3:self.days + 3],
            binary.strip().split(' ')[4:self.days + 4],
            binary.strip().split(' ')[5:self.days + 5],
            binary.strip().split(' ')[6:self.days + 6]
        ]

    def count_binary(self, binary):
        arr = []
        for b in binary:
            arr.append(b.count('1'))
        return arr

days = str(sys.stdin.read())
a = A(days)
binary = a.get_binary()
binary = a.get_binary_array(binary)
binary = a.count_binary(binary)
print(str(min(binary)) + " " + str(max(binary)))