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 ai. 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?

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 ai (0?≤?ai?≤?l). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.

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.

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：一个长l的街道，在一些指定点上会有一些灯笼，问灯笼最小照亮半径为多少可以照亮整条街。

处理下，相邻元素最大值，注意一下0 or L处没有灯笼的情况。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <set>
#include <limits>
#include <map>

using namespace std;

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define F(i, n) for(int (i)=0;(i)<(n);++(i))
#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))
#define MEM0(addr) memset((addr), 0, sizeof((addr)))

#define PI 3.1415926535897932384626433832795
#define HALF_PI 1.5707963267948966192313216916398

#define MAXN 100000
#define MAXM 10000
#define MOD 1000000007

typedef long long LL;

const double maxdouble = numeric_limits<double>::max();
const double eps = 1e-10;
const int INF = 0x7FFFFFFF;



int a[1005];
int main()
{
    int n,l;
    while(cin>>n>>l)
    {
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        sort(a,a+n);
        int ma = -1;
        for(int i=1;i<n;i++)
        {
            int tmp = a[i] - a[i-1];
            ma = max(tmp,ma);
        }
        int x = 2*(a[0] - 0);
        int y = 2*(l - a[n-1]);
        ma = max(ma,x);
        ma = max(ma,y);
        double ans = (double) ma/2;
        printf("%.10lf\n",ans);
    }
}