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
  
  
  factarr=[1]
  maxi=2*(10**5)
  res=1
  for i in range(1,maxi+5):
      res=(res*i)%mod
      factarr.append(res)
  
  #a^n % mod in logn
  def power(a,n):
      r=1
      while(n):
          if(n%2==1):
              r=(r*a)%mod
          n=n//2
          a=(a*a)%mod
      return r
  
  # Function to return nCr % p in O(1) time
  # n C r = n!*inverse(r!)*inverse((n-r)!)
  #inverse(x)%m=pow(x,m-2)%m
  def ncr(n, r, p):
      if(r>n):
          return 0
      res=factarr[n]
      res=(res*power(factarr[r],p-2))%p
      res=(res*power(factarr[n-r],p-2))%p
      return res
  
  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
      if(l[-1]==l[-2]):
          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*ncr(n-count-1,i-count-1,mod))%mod  #ncr mod p in O(1) using precomputing
              ans=(ans-tot+mod)%mod
          print(ans)
  
      