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>=)
     {
       s1=(n/)*;
       if(n%==)
       s1+=;
     }
     else
     {
      if(n==)
      s1=;
      else
      s1=;
     }
     int s2=;
     if(n->=)
     {
       s2+=((n-)/*);
       if((n-)%==)
          s2+=;
     }
     else
     {
      if(n-==)
      s2+=;
      else
      s2+=;
     }
     if(n>)
     printf("%d %d\n",s1,s2);
     if(n==)
     printf("0 1\n");
     if(n==)
     printf("0 2\n");
 }
     return ;
 }