Codeforces Round #350 (Div. 2) A

A. Holidays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意： 假设一年有n天 问一年内的休息天的数量（一周7天 5+2） 的最小值与最大值

题解：最小值---从周一开始算

         最大值---从周六开始算

   比赛的时候这题都gg了 没处理好 菜

#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<map> 
#define ll __int64
#define pi acos(-1.0)
using namespace std;
int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
    int s1;
    if(n>=7)
    {
      s1=(n/7)*2;
      if(n%7==6)
      s1+=1; 
    }
    else
    {
     if(n==6)
     s1=1;
     else
     s1=0;
    }
    int s2=2;
    if(n-2>=7)
    { 
      s2+=((n-2)/7*2);
      if((n-2)%7==6)
         s2+=1;
    }
    else
    {
     if(n-2==6)
     s2+=1;
     else
     s2+=0;
    }
    if(n>2)
    printf("%d %d
",s1,s2);
    if(n==1)
    printf("0 1
");
    if(n==2)
    printf("0 2
");
}
    return 0; 
}