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 gcd(a, b):
      if a < b:
          b, a = a, b
      while a * b > 0:
          a %= b
          b, a = a, b
      return a
  
  
  def candy(k, c):
      if gcd(k, c) > 1:
          return -1
      s = [1, 0]
      r = [c, k]
      q = [0]
      i = 2
      while r[i - 1] != 0:
          q.append(r[i - 2] // r[i - 1])
          s.append(s[i - 2] - s[i - 1] * q[i - 1])
          r.append(r[i - 2] - r[i - 1] * q[i - 1])
          i += 1
      return s[i - 2] % k
  
  
  def sex(n, a):
      m = max(a)
      idx = a.index(m)
      okay = False
      for i in range(n):
          if i != idx and m - a[i] <= 1:
              okay = True
              break
      if not okay:
          return 0
      okay = False
      for i in range(idx + 1, n):
          if a[i] == m:
              okay = True
              break
      if okay:
          return fac[n]
      tot = 0
      for i in range(n):
          if a[i] == m - 1:
              tot += 1
      return (fac[n] - fac[n] * candy(p, tot + 1)) % p
  
  
  p = 998244353
  fac = [1] * 200001
  for i in range(2, 200001):
      fac[i] = fac[i - 1] * i
      fac[i] %= p
  t = int(input())
  for _ in range(t):
      n = int(input())
      a = list(map(int, input().split()))
      print(sex(n, a))