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
  input = sys.stdin.readline
  from collections import *
  
  MAX = 2*10**5+100
  MOD = 998244353
  fact = [0]*MAX #fact[i]: i!
  inv = [0]*MAX #inv[i]: iの逆元
  finv = [0]*MAX #finv[i]: i!の逆元
  fact[0] = 1
  fact[1] = 1
  finv[0] = 1
  finv[1] = 1
  inv[1] = 1
      
  for i in range(2, MAX):
      fact[i] = fact[i-1]*i%MOD
      inv[i] = MOD-inv[MOD%i]*(MOD//i)%MOD
      finv[i] = finv[i-1]*inv[i]%MOD
  
  def C(n, r):
      if n<r:
          return 0
      if n<0 or r<0:
          return 0
      return fact[n]*(finv[r]*finv[n-r]%MOD)%MOD
  
  for _ in range(int(input())):
      n = int(input())
      a = list(map(int, input().split()))
      a.sort()
      
      if a[-1]-a[-2]>=2:
          print(0)
          continue
     
      if a[-1]==a[-2]:
          print(fact[n])
          continue
      
      c = a.count(a[-2])
      print((fact[n]-C(n, c+1)*fact[n-c-1]*fact[c]%MOD)%MOD)