CodeForces 709B Checkpoints

B. Checkpoints

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, ..., xn. Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.

Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.

Input

The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000,  - 1 000 000 ≤ a ≤ 1 000 000) — the number of checkpoints and Vasya's starting position respectively.

The second line contains n integers x1, x2, ..., xn ( - 1 000 000 ≤ xi ≤ 1 000 000) — coordinates of the checkpoints.

Output

Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.

Examples

input

3 10
1 7 12

output

7

input

2 0
11 -10

output

10

input

5 0
0 0 1000 0 0

output

0

将n个点排序之后

n个点要走n-1个点，那么最后一个点肯定呢是边界两个点之一，分两种情况讨论

每种情况，可以根据起点和n-1个点两端的位置关系分类

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
const int maxn=1e5;
int n,start;
int a[maxn+5];
int main()
{
    scanf("%d%d",&n,&start);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    if(n==1)
    {
        printf("0
");
        return 0;
    }
    int res1,res2;
    sort(a+1,a+n+1);
    if(start<=a[2])
        res1=abs(a[n]-start);
    else if(start>=a[n])
        res1=abs(a[2]-start);
    else
        res1=min(abs(a[2]-start)+abs(a[2]-a[n]),abs(a[n]-start)+abs(a[2]-a[n]));

    if(start<=a[1])
        res2=abs(a[n-1]-start);
    else if(start>=a[n-1])
        res2=abs(a[1]-start);
    else
        res2=min(abs(a[1]-start)+abs(a[1]-a[n-1]),abs(a[n-1]-start)+abs(a[1]-a[n-1]));
    printf("%d
",min(res1,res2));
    return 0;
}