Codeforces 723 A. The New Year: Meeting Friends

A. The New Year: Meeting Friends

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are three friend living on the straight line Ox in Lineland. The first friend lives at the point x₁, the second friend lives at the point x₂, and the third friend lives at the point x₃. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?

It's guaranteed that the optimal answer is always integer.

Input

The first line of the input contains three distinct integers x₁, x₂ and x₃ (1 ≤ x₁, x₂, x₃ ≤ 100) — the coordinates of the houses of the first, the second and the third friends respectively.

Output

Print one integer — the minimum total distance the friends need to travel in order to meet together.

Examples

Input

7 1 4

Output

6

Input

30 20 10

Output

20

Note

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

 

代码1:

#include<stdio.h>
int main(){
    int a,b,c,t,dis;
    while(~scanf("%d %d %d",&a,&b,&c)){
       if(a<b)swap(a,b);if(a<c)swap(a,c);if(b<c)swap(b,c);
       dis=a-c;
       printf("%d
",dis);
    }
   return 0;
}

代码2:

#include<stdio.h>
#include<math.h>
int main(){
    int a,b,c,dis,m;
    while(~scanf("%d %d %d",&a,&b,&c)){
        m=1000;
        for(int i=1;i<=100;i++){
        　　dis=fabs(a-i)+fabs(b-i)+fabs(c-i);
       　　 if(m>dis)
            　　m=dis;
        }
        printf("%d
",m);
    }
    return 0;
}