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 sys
  read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
  import bisect,string,math,time,functools,random,fractions
  from heapq import heappush,heappop,heapify
  from collections import deque,defaultdict,Counter
  from itertools import permutations,combinations,groupby
  rep=range;R=range
  def I():return int(input())
  def LI():return [int(i) for i in input().split()]
  def LI_():return [int(i)-1 for i in input().split()]
  def S_():return input()
  def IS():return input().split()
  def LS():return [i for i in input().split()]
  def NI(n):return [int(input()) for i in range(n)]
  def NI_(n):return [int(input())-1 for i in range(n)]
  def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
  def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
  def StoLI():return [ord(i)-97 for i in input()]
  def ItoS(n):return chr(n+97)
  def LtoS(ls):return ''.join([chr(i+97) for i in ls])
  def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
  def RI(a=1,b=10):return random.randint(a,b)
  def INP():
      N=10**6
      n=random.randint(1,N)
      a=[random.randint(1,n) for i in range(n)]
      return n,a
  def Rtest(T):
      case,err=0,0
      for i in range(T):
          inp=INP()
          a1=naive(*inp)
          a2=solve(*inp)
          if a1!=a2:
              print(inp)
              print('naive',a1)
              print('solve',a2)
              err+=1
          case+=1
      print('Tested',case,'case with',err,'errors')
  def GI(V,E,ls=None,Directed=False,index=1):
      org_inp=[];g=[[] for i in range(V)]
      FromStdin=True if ls==None else False
      for i in range(E):
          if FromStdin:
              inp=LI()
              org_inp.append(inp)
          else:
              inp=ls[i]
          if len(inp)==2:a,b=inp;c=1
          else:a,b,c=inp
          if index==1:a-=1;b-=1
          aa=(a,c);bb=(b,c);g[a].append(bb)
          if not Directed:g[b].append(aa)
      return g,org_inp
  def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
      #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
      mp=[boundary]*(w+2);found={}
      for i in R(h):
          s=input()
          for char in search:
              if char in s:
                  found[char]=((i+1)*(w+2)+s.index(char)+1)
                  mp_def[char]=mp_def[replacement_of_found]
          mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
      mp+=[boundary]*(w+2)
      return h+2,w+2,mp,found
  def TI(n):return GI(n,n-1)
  def accum(ls):
      rt=[0]
      for i in ls:rt+=[rt[-1]+i]
      return rt
  def bit_combination(n,base=2):
      rt=[]
      for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
      return rt
  def gcd(x,y):
      if y==0:return x
      if x%y==0:return y
      while x%y!=0:x,y=y,x%y
      return y
  def YN(x):print(['NO','YES'][x])
  def Yn(x):print(['No','Yes'][x])
  def show(*inp,end='\n'):
      if show_flg:print(*inp,end=end)
  
  mo=10**9+7
  #mo=998244353
  inf=1<<63
  FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
  alp=[chr(ord('a')+i)for i in range(26)]
  #sys.setrecursionlimit(10**7)
  
  
  show_flg=False
  show_flg=True
  
  ans=0
  
  mo=998244353
  f=[1]
  n=2*10**5+1
  for i in range(1,n+1):
      f+=f[-1]*i%mo,
      
  for _ in range(I()):
      ans=0
      n=I()
      a=LI()
      a.sort()
      if a[0]==a[-1]:
          ans=f[n]
      else:
          m=a.count(a[-1])
          s=a.count(a[-1]-1)
          if m>1:
              ans=f[n]
          else:
              if a[-1]-1 not in a:
                  ans=0
              else:
                  ans=f[n]*s*pow(m+s,mo-2,mo)%mo
      print(ans%mo)
  
  
  