description: n people gathered to hold a jury meeting of the upcoming competition, the i-th member of the jury came up with a_i tasks, which they want to share with each other.
  
  First, the jury decides on the order which they will follow while describing the tasks. Let that be a permutation p of numbers from 1 to n (an array of size n where each integer from 1 to n occurs exactly once).
  
  Then the discussion goes as follows:
  
    * If a jury member p_1 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * If a jury member p_2 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * ... 
    * If a jury member p_n has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * If there are still members with tasks left, then the process repeats from the start. Otherwise, the discussion ends. 
  
  
  
  A permutation p is nice if none of the jury members tell two or more of their own tasks in a row. 
  
  Count the number of nice permutations. The answer may be really large, so print it modulo 998 244 353.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first line of the test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — number of jury members.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of problems that the i-th member of the jury came up with.
  
  The sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print one integer — the number of nice permutations, taken modulo 998 244 353.
  
  Example
  
  Input
  
  
  4
  2
  1 2
  3
  5 5 5
  4
  1 3 3 7
  6
  3 4 2 1 3 3
  
  
  Output
  
  
  1
  6
  0
  540
  
  Note
  
  Explanation of the first test case from the example:
  
  There are two possible permutations, p = [1, 2] and p = [2, 1]. For p = [1, 2], the process is the following:
  
    1. the first jury member tells a task; 
    2. the second jury member tells a task; 
    3. the first jury member doesn't have any tasks left to tell, so they are skipped; 
    4. the second jury member tells a task. 
  
  
  
  So, the second jury member has told two tasks in a row (in succession), so the permutation is not nice.
  
  For p = [2, 1], the process is the following:
  
    1. the second jury member tells a task; 
    2. the first jury member tells a task; 
    3. the second jury member tells a task. 
  
  
  
  So, this permutation is nice.
solution: import sys
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  inf = 10**16
  # md = 10**9+7
  md = 998244353
  
  def nCr(com_n, com_r):
      if com_r < 0: return 0
      if com_n < com_r: return 0
      return fac[com_n]*ifac[com_r]%md*ifac[com_n-com_r]%md
  
  # 準備
  n_max = 200005
  fac = [1]
  for i in range(1, n_max+1): fac.append(fac[-1]*i%md)
  ifac = [1]*(n_max+1)
  ifac[n_max] = pow(fac[n_max], md-2, md)
  for i in range(n_max-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%md
  
  def fs(aa):
      f = s = -1
      for a in aa:
          if a >= f: f, s = a, f
          elif a > s: s = a
      return f, s
  
  def solve():
      n = II()
      aa = LI()
      f, s = fs(aa)
      if f == s:
          print(fac[n])
      elif f-s > 1:
          print(0)
      else:
          c=aa.count(s)
          ans=(fac[n]-fac[n]*pow(c+1,md-2,md)%md)%md
          print(ans)
  
  for testcase in range(II()):
      solve()
  