Codeforces 723A The New Year: Meeting Friends（题意有偏差的水题）

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 pointx₁, the second friend lives at the pointx₂, and the third friend lives at the pointx₃. 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 integersx₁,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 of3 (from the point7 to the point4), the second friend also has to travel the distance of3 (from the point1 to the point4), while the third friend should not go anywhere because he lives at the point4.

思路：

很水的一题。。但是我一开始题意理解错了。。。给你这三个点，算他们走到一起的最短距离。我一开始以为能在这区间的所有点走呢。。结果是只能到达这三个点中的其中一个。。。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath> 
#include<algorithm> 
using namespace std;


int main()
{
	int x,y,z;
	while(scanf("%d%d%d",&x,&y,&z)!=EOF)
	{
		int sum1,sum2,sum3; 
		sum1=abs(x-y)+abs(x-z);
		sum2=abs(x-y)+abs(y-z);
		sum3=abs(x-z)+abs(y-z);
		int sum=min(sum1,sum2);
		sum=min(sum,sum3);
		printf("%d\n",sum);
	}
	return 0;
}