Codeforces Round #350 (Div. 2) A

Description

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

output

4 4

input

output

0 2
给我一个天数，问最少能过几个假期，最多能过几个假期~
分情况看余数~
讨论就好了

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<stack>
#include<math.h>
#include<map>
#include<sstream>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define LLL l,m,rt<<1
#define RRR r,m+1,rt<<1|1
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n==1)
    {
        cout<<"0"<<" "<<"1"<<endl;
    }
    else if(2<=n&&n<=5)
    {
        cout<<"0"<<" "<<"2"<<endl;
    }
    else if(n==6)
    {
        cout<<"1"<<" "<<"2"<<endl;
    }
    else
    {
        int pot;
        int pos;
        pot=n/7*2;
        pos=n/7*2;
        if(n%7>=0&&n%7<=5)
        {
            cout<<pot<<" ";
        }
        else
        {
            cout<<pot+1<<" ";
        }
        if(n%7==1)
        {
            cout<<pos+1<<endl;
        }
        else if(n%7==0)
        {
            cout<<pos<<endl;
        }
        else
        {
            cout<<pos+2<<endl;
        }
    }
    return 0 ;
}