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: def solve(n,tasks):
      mod_val = 998244353
      max1 = 0
      max2 = 0
      for t in tasks:
          if t > max1:
              max2 = max1
              max1 = t
          elif t > max2:
              max2 = t
      if max1 > max2 + 1:
          return 0
      if max1 == max2+1:
          count2 = tasks.count(max2)
          tot = get_factorial(n, mod_val)
          for i in range(n):
              if i >= count2:
                  tmp = (choose(n-count2-1,i-count2, mod_val) *get_factorial(n-i-1, mod_val))%mod_val
                  tmp = (tmp*get_factorial(i, mod_val))%mod_val
                  tot = (tot - tmp)%mod_val
                  # print(tot)
          return tot
      res = get_factorial(n, mod_val)
  
      # print(tasks)
      return res
  
  fac_cache={}
  inv_fac_cache = {}
  mod_val = 998244353
  
  def invert_mod(i, mod_val):
      return pow(i, mod_val-2, mod_val)
  
  def choose(N,k, mod_val):
      if k == 0:
          return 1
      if k == N:
          return 1
      if k == 1:
          return N
      if k == (N-1):
          return N
      n_fac = get_factorial(N, mod_val)
      nk_inv_fac = get_inv_factorial(N-k, mod_val)
      k_inv_fac = get_inv_factorial(k, mod_val)
      res = (n_fac * nk_inv_fac) % mod_val
      res = (res * k_inv_fac) % mod_val
      return res
  
  # if you run the biggest factorial first then all are cached
  def get_factorial(N, mod_val):
      cached_res = fac_cache.get(N, None)
      if cached_res:
          return cached_res
      res = 1
      for i in range(2, N+1):
          res = (res * i) %mod_val
          fac_cache[i] = res
      return res
  
  
  def get_inv_factorial(k, mod_val):
      cached_res = inv_fac_cache.get(k, None)
      if cached_res:
          return cached_res
      res = 1
      for i in range(2, k+1):
          res = (res * invert_mod(i, mod_val)) % mod_val
          inv_fac_cache[i] = res
      return res
  
  import os
  import io
  import time
  if __name__ == "__main__":
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
      T = int(input().decode().strip())
      for t in range(T):
          n = int(input().decode().strip())
          tasks = [int(x) for x in input().decode().strip().split(" ")]
          res = solve(n,tasks)
          print(res)