There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T. 2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn't equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can't coincidently at the same position.

There are several test cases. There is a number T ( T <= 50 ) on the first line which shows the number of test cases. For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line. On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.

For each test cases , output a real number shows the answser. Please output three digit after the decimal point.

3
3
1 2 3
3
1 2 4
4
1 9 100 10

1.000
2.000
8.000

 #include <iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<cstring>

 #include<algorithm>

 #include<vector>

 #include<map>

 #include<string>

 #include<cmath>

 #include<queue>

 #define N 1005

 #define MAXN 1000005

 #define M 200

 #define mod 1000000007

 #define ll long long

 using namespace std;

 int n,m;

 int T;

 ll a[N],b[N];

 ll te;

 ll ans;

 bool cmp(ll x,ll y)

 {

     return x>y;

 }

 int isok(ll v)

 {

     ll now=;

     for(int i=;i<=n-;i++){

        // printf(" %I64d %I64d %I64d now=%I64d v=%I64d\n",a[i+1],a[i],a[i-1],now,v);

        // printf(" %d\n",(a[i]-a[i-1]-now) <=v);

         if( (a[i]-a[i-]-now) >=v){

             now=;

         }

         else{

             if( (a[i+]-a[i]) ==v){

                 now=;

                 i++;

             }

             else if( (a[i+]-a[i]) >v){

                 now=v;

             }

             else return ;

         }

     }

     return ;

 }

 int main()

 {

     int i,j,k;

     //freopen("data.txt","r",stdin);

     scanf("%d",&T);

     //while(scanf("%d",&n)!=EOF)

     while(T--)

    // for(cnt=1;cnt<=T;cnt++)

     {

        scanf("%d",&n);

        m=;

        for(i=;i<=n;i++){

             scanf("%I64d",&a[i]);

             a[i]*=;

        }

        sort(a+,a++n);

        //for(i=1;i<=n;i++)printf(" %I64d\n",a[i]);

        for(i=;i<n;i++){

             te=a[i+]-a[i];

             b[m]=te;m++;

             te/=;

             b[m]=te;m++;

        }

        sort(b,b+m,cmp);

        //for(i=0;i<m;i++)printf(" %I64d\n",b[i]);

        for(j=;j<m;j++){

             if(isok(b[j])==){

                 printf("%.3f\n",(double)b[j]/);

                 break;

             }

        }

     }

     return ;

 }