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: 
  
  mod=998244353
   
  def make_mod_mul(mod = 10**9 + 7):
      fmod_inv = 1.0 / mod
      def mod_mul(a, b, c=0):
          res = a * b +  c - mod *  int(fmod_inv * a * b + fmod_inv * c)
          if res >= mod:
              return res - mod
          elif res < 0:
              return res + mod
          else:
              return res
      return mod_mul
   
  mod_mul=make_mod_mul(mod)
  def mod_pow(x, y):
      if y == 0:
          return 1
      res = 1
      while y > 1:
          if y & 1 == 1:
              res = mod_mul(res,x)
          x = mod_mul(x, x)
          y >>= 1
      return mod_mul(res, x)
   
  MAX=200005
  fact=[1]
  for zz in range(1,MAX+1):
      fact.append(mod_mul(fact[-1],zz))
  def nCr(n,r): #choose, MOD is prime
      num=fact[n]
      den=mod_mul(fact[r],fact[n-r])
      return (num*pow(den,mod-2, mod))%mod
   
      
  t = int(input())
  for _ in range(t):
      n = int(input())
      a = list(map(int, input().split()))
      a = sorted(a)
      ans = 'yes'
      m = max(a)
      mm = a[0]
      for i in range(1, len(a)):
          if a[i] > mm and a[i] != m:
              mm = a[i]
  
      if a[-1] == a[-2]:
          print(fact[n])
      elif m-mm>1:
          print(0)
      elif m == mm:
          print(fact[n])
      elif a[-1] == a[-2] and n == 2:
          print(2)
      
      else:
          k = 0
          
          for i in range(len(a)):
              if a[i] == mm:
                  k += 1
            
          ans = fact[n]
          for i in range(k, n):
              #print(i, ans,nCr(i, i-k) * fact[k])
              ans = (ans + mod - nCr(i, i-k) * fact[k]*fact[n-k-1])%mod
              
          if n == 2:
              print(1)
          else:
              print(ans)