Description

Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point l. Then the i-th lantern is at the point a_i. The lantern lights all points of the street that are at the distance of at most d from it, where d is some positive number, common for all lanterns.

Vanya wonders: what is the minimum light radius d should the lanterns have to light the whole street?

Input

The first line contains two integers n, l (1 ≤ n ≤ 1000, 1 ≤ l ≤ 10⁹) — the number of lanterns and the length of the street respectively.

The next line contains n integers a_i (0 ≤ a_i ≤ l). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.

Output

Print the minimum light radius d, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10^- 9.

Sample Input

Input

7 15 15 5 3 7 9 14 0

Output

2.5000000000

Input

2 5 2 5

Output

2.0000000000

Hint

Consider the second sample. At d = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit.

 

 

分析：

       只要找到灯笼与灯笼之间最远的距离就好了,这样所有的灯笼着凉整条路就都覆盖了。

 

 

源代码：

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 #include
           
             6 using namespace std; 7 #define maxn 1000+10 8 int arr[maxn]; 9 long long l;10 int main()11 {12     int n;13     while (cin >> n >> l)14     {   15         double r0,r2,r;16         double d=0.0;17         for (int i = 0; i < n; i++)18         {19             cin >> arr[i];20         }21         sort(arr, arr + n);22         r0 =arr[0] - 0;                              //左端点的半径23         r2 =max(r0,(double) l - arr[n - 1]);            //从左端点跟右端点找一个最大的半径24         for (int j = 0; j < n; j++)25         {26             d = max(d, (double)arr[j + 1] - arr[j]);       //找灯笼间距离最远的区间27         }28         r =d / 2.0;                     //灯笼间最远距离的一半，就是半径29         r =max(r2, r);                   //与前面的端点半径比较，找最大的30         printf("%.10f\n", r);31     }32     return 0;33 }