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 heapq,math
  from collections import defaultdict,deque
  from os import getcwd
  from itertools import permutations
  #from functools import cmp_to_key
  
  import sys, os.path
  #sys.setrecursionlimit(10000000)
  
  if(os.path.exists('C:/Users/Dhanush/Desktop/cp/input.txt')):
      sys.stdout = open('C:/Users/Dhanush/Desktop/cp/output.txt', 'w')
      sys.stdin = open('C:/Users/Dhanush/Desktop/cp/input.txt', 'r')
  
  input=sys.stdin.readline
  mod=998244353
  
  
  
  N = 200001
   
  # array to store inverse of 1 to N
  factorialNumInverse = [None] * (N + 1)
   
  # array to precompute inverse of 1! to N!
  naturalNumInverse = [None] * (N + 1)
   
  # array to store factorial of
  # first N numbers
  fact = [None] * (N + 1)
   
  # Function to precompute inverse of numbers
  def InverseofNumber(p):
      naturalNumInverse[0] = naturalNumInverse[1] = 1
      for i in range(2, N + 1, 1):
          naturalNumInverse[i] = (naturalNumInverse[p % i] *
                                     (p - int(p / i)) % p)
   
  # Function to precompute inverse
  # of factorials
  def InverseofFactorial(p):
      factorialNumInverse[0] = factorialNumInverse[1] = 1
   
      # precompute inverse of natural numbers
      for i in range(2, N + 1, 1):
          factorialNumInverse[i] = (naturalNumInverse[i] *
                                    factorialNumInverse[i - 1]) % p
   
  # Function to calculate factorial of 1 to N
  def factorial(p):
      fact[0] = 1
   
      # precompute factorials
      for i in range(1, N + 1):
          fact[i] = (fact[i - 1] * i) % p
   
  # Function to return nCr % p in O(1) time
  def Binomial(N, R, p):
       
      # n C r = n!*inverse(r!)*inverse((n-r)!)
      ans = ((fact[N] * factorialNumInverse[R])% p *
                        factorialNumInverse[N - R])% p
      return ans
  
  def isnice(l):
      i=0
      prev=-1
      cnt=n
      while(cnt>0):
          if(l[i]==1):
              if(prev==i):
                  return False
              l[i]=0
              cnt-=1
              prev=i
          elif(l[i]>1):
              if(prev==i):
                  return False
              l[i]-=1
              prev=i
          else:
              if(prev==i):
                  return True
          i=(i+1)%n
  
  
      return True
  
  factarr=[1]
  maxi=2*(10**5)
  p = mod
  InverseofNumber(p)
  InverseofFactorial(p)
  factorial(p)
  res=1
  for i in range(1,maxi+5):
      res=(res*i)%mod
      factarr.append(res)
  
  
  def ncr(n, r, p):
      num = den = 1
      for i in range(r):
          num = (num * (n - i)) % p
          den = (den * (i + 1)) % p
      return (num * pow(den,
              p - 2, p)) % p
  
  tc=int(input())
  for _ in range(tc):
      n=int(input())
      l=sorted(list(map(int,input().split())))
      d=defaultdict(int)
      for i in l:
          d[i]+=1
          '''
      perm=list(permutations(l))
      print(len(perm),perm)
      c=0
      for row in perm:
          if(isnice(list(row))):
              c+=1
              print(row)
          else:
              print('not true',row)
      print(c)
  
  
          '''
      
      if(l[-1]==l[-2]):
          #ans is n!
          print(factarr[n])
      elif(l[-1]>l[-2]+1):
          print(0)
      else:
          prev=l[-2]
          count=d[prev]
          ans=factarr[n]
          for i in range(count+1,n+1):
              c1=factarr[i-1]
              c2=factarr[n-i]
              tot=(c1*c2)%mod
              tot=(tot*Binomial(n-count-1,i-count-1,mod))%mod
              ans=(ans-tot+mod)%mod
          print(ans)
  
      