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.path
  from collections import *
  import math
  import bisect
  import heapq as hq
  from fractions import Fraction
  from random import randint
  
  if (os.path.exists('input.txt')):
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
  
  
  ##########################################################
  
  input = sys.stdin.readline
  mod = 998244353
  
  fact = [1] * (200005)
  
  for i in range(1,len(fact)):
      fact[i] = (fact[i - 1] * i) % mod
  
  def power(x, y, m):
   
      if (y == 0):
          return 1
   
      p = power(x, y // 2, m) % m
      p = (p * p) % m
   
      if(y % 2 == 0):
          return p
      else:
          return ((x * p) % m)
  
  def modInverse(a):
      return power(a, mod - 2, mod)
  
  def solve():
      n = int(input())
      arr = [int(x) for x in input().split()]
      arr.sort()
      if arr[-1] - arr[-2] > 1:
          print(0)
          return
      if arr[-1] == arr[-2]:
          print(fact[n])
          return
  
      res = 0
      x = arr.index(arr[-2])
  
      num = 1
      den = 1
      res = 0
  
      for i in range(x):
          num = (num * (x - i)) % mod
          den = (den * (i + 1)) % mod
          y = 1
          y = (y * fact[n - (i + 2)]) % mod
          y = (y * fact[i + 1]) % mod
          y = (y * num) % mod
          y = (y * modInverse(den)) % mod
          res = (res + y) % mod
  
      print((fact[n] - (fact[n - 1] + res) % mod + mod) % mod)
  
  t = int(input())
  while t:
      t -= 1
      solve()
  
  ##########################################################
  