Miaomiao's Geometry

Accepts: 21
Submissions: 1198
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
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.
Input
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.
Output
For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
Sample Input
3
3
1 2 3
3
1 2 4
4
1 9 100 10
Sample Output
1.000
2.000
8.000
Hint
For the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.