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
  import os
  from io import BytesIO
  
  if sys.version_info[0] < 3:
      sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
  
  f = sys.stdin
  if os.environ.get('USER') == "loic":
      f = open("data.in")
  
  line = lambda: f.readline().strip('\r\n').split()
  
  def write(w):
      sys.stdout.write(w)
      sys.stdout.write("\n")
  
  def factorial_modulo(limit):
      fact = [1]
      for i in range(1, limit+1):
          fact.append(fact[-1] * i % M)
      return fact
  
  def inv_modulo_all(limit):
      mod_inv = [0,1]
      for i in range(2, limit+1):
          mod_inv.append(M - (M // i) * mod_inv[M % i] % M)
      return mod_inv
  
  def factorial_modulo_inverse(limit):
      mod_inv = inv_modulo_all(limit)
      fact_inv = [1,1]
      for i in range(2, limit+1):
          fact_inv.append(fact_inv[-1] * mod_inv[i] % M)
      return fact_inv
  
  def solve():
      d = sorted(D,reverse=True)
      a = d[0]
      b = d[1]
      c = 1
      for i in range(2,N):
          if d[i] == b:
              c += 1
          else:
              break
      
      if a - b > 1:
          return str(0)
      if a==b:
          return str(FACT[N])
      
      res = c * FACT[N-1]
      res %= M
      
      for i in range(c,N-1):
          tmp = FACT[i] * FACT_INV[i-c] * FACT[N-1-c]
          res += FACT[N-1] - tmp
          res %= M
      
      return str(res)
  
  M = 998244353
  Z = 200000
  FACT = factorial_modulo(Z)
  FACT_INV = factorial_modulo_inverse(Z)
  
  T = int(line()[0])
  for test in range(1,T+1):
      N = int(line()[0])
      D = list(map(int,line()))
      write(solve())
      
  f.close()