description: Ezzat has an array of n integers (maybe negative). He wants to split it into two non-empty subsequences a and b, such that every element from the array belongs to exactly one subsequence, and the value of f(a) + f(b) is the maximum possible value, where f(x) is the average of the subsequence x. 
  
  A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.
  
  The average of a subsequence is the sum of the numbers of this subsequence divided by the size of the subsequence.
  
  For example, the average of [1,5,6] is (1+5+6)/3 = 12/3 = 4, so f([1,5,6]) = 4.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^3)— the number of test cases. Each test case consists of two lines.
  
  The first line contains a single integer n (2 ≤ n ≤ 10^5).
  
  The second line contains n integers a_1, a_2, …, a_n (-10^9 ≤ a_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 3⋅10^5.
  
  Output
  
  For each test case, print a single value — the maximum value that Ezzat can achieve.
  
  Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
  
  Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
  
  Example
  
  Input
  
  
  4
  3
  3 1 2
  3
  -7 -6 -6
  3
  2 2 2
  4
  17 3 5 -3
  
  
  Output
  
  
  4.500000000
  -12.500000000
  4.000000000
  18.666666667
  
  Note
  
  In the first test case, the array is [3, 1, 2]. These are all the possible ways to split this array: 
  
    * a = [3], b = [1,2], so the value of f(a) + f(b) = 3 + 1.5 = 4.5. 
    * a = [3,1], b = [2], so the value of f(a) + f(b) = 2 + 2 = 4. 
    * a = [3,2], b = [1], so the value of f(a) + f(b) = 2.5 + 1 = 3.5. 
  
  Therefore, the maximum possible value 4.5.
  
  In the second test case, the array is [-7, -6, -6]. These are all the possible ways to split this array: 
  
    * a = [-7], b = [-6,-6], so the value of f(a) + f(b) = (-7) + (-6) = -13. 
    * a = [-7,-6], b = [-6], so the value of f(a) + f(b) = (-6.5) + (-6) = -12.5. 
  
  Therefore, the maximum possible value -12.5.
solution: from collections import Counter, defaultdict, deque
  import bisect
  from sys import stdin, stdout
  from itertools import repeat, permutations
  import math
  import heapq
  
  MOD = 998244353
  def inp(force_list=False):
      re = map(int, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
  
  def inpf(force_list=False):
      re = map(float, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
  
  def inst():
      return raw_input().strip()
  
  def gcd(x, y):
     while(y):
         x, y = y, x % y
     return x
  
  def lcm(x, y):
      return x*y/(gcd(x,y))
  
  def qmod(a, b, mod=MOD):
      res = 1
      while b:
          if b&1:
              res = (res*a)%mod
          b >>= 1
          a = (a*a)%mod
          # print b
      return res
  
  def inv(a):
      return qmod(a, MOD-2)
  
  def query(num, k):
      print tmk(num, k)
      stdout.flush()
      return inp()
  
  class RMQ(object):
      def __init__(self, da):
          self.n = len(da)
          self.da = [[0 for i in range(32)] for i in range(self.n)]
          for i in range(self.n):
              self.da[i][0] = da[i]
          for j in range(1, 32):
              for i in range(self.n):
                  self.da[i][j] = gcd(self.da[i][j-1], self.da[(i+(1<<(j-1)))%self.n][j-1])
  
      def query(self, l, r):
          j = int(math.log(r-l+1, 2))
          return gcd(self.da[l][j], self.da[(r-(1<<j)+1)%self.n][j])
  
  
  def getsum(BITTree,i):
      if i<0:
          return 0
      s = 0
      i = i+1
      while i > 0:
          s += BITTree[i]
          i -= i & (-i)
      return s
  
  def updatebit(BITTree , n , i ,v):
      i += 1
      while i <= n:
          BITTree[i] += v
          i += i & (-i)
  
  def construct(arr, n):
  
      BITTree = [0]*(n+1)
      for i in range(n):
          updatebit(BITTree, n, i, arr[i])
      return BITTree
  
  def GV(num, s, p):
      num = num**s - (num%p)**s
      re = 0
      while num and num%p==0:
          re+=1
          num/=p
      return re
  
  def my_main():
      kase = inp()
      pans = []
      for ka in range(1, kase+1):
          # print "Case #%d:" % ka,
          n = inp()
          da = inp()
          da.sort()
          pans.append(str( da[-1]/float(1) + sum(da[:-1])/float(n-1)))
      print '\n'.join(pans)
  
  
  my_main()
  