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: """
  ID: happyn61
  LANG: PYTHON3
  PROB: loan
  """
  from itertools import product
  import itertools
  import math
  #from collections import defaultdict
  import sys
  import heapq
  from collections import deque
  MOD=1000000000007
  #fin = open ('loan.in', 'r')
  #fout = open ('loan.out', 'w')
  #print(dic["4734"])
  def find(parent,i):
  
  
      if parent[i] != i: 
          parent[i]=find(parent,parent[i]) 
      return parent[i] 
  
          # A utility function to do union of two subsets 
  def union(parent,rank,xx,yy): 
      x=find(parent,xx)
      y=find(parent,yy)
      if rank[x]>rank[y]:
          parent[y]=x
      elif rank[y]>rank[x]:
          parent[x]=y
      else:
          parent[y]=x
          rank[x]+=1
  ans=0
  #NK=sys.stdin.readline().strip().split()
  K=int(sys.stdin.readline().strip())
  #N=int(NK[0])
  #K=int(NK[1])
  #M=int(NK[2])
  #ol=list(map(int,sys.stdin.readline().strip().split()))
  #d={0:0,1:0}
  
  x=0
  y=0
  
  #d={"N":(0,1),"S":(0,-1),"W":(-1,0),"E":(1,0)}
  for _ in range(K):
      n=int(sys.stdin.readline().strip())
      s=list(map(int,sys.stdin.readline().strip().split()))
      #print(s)
      s.sort(reverse=True)
      if s[0]==s[1]:
          print(math.factorial(n)%998244353)
      else:
          if s[0]-s[1]>1:
              print(0)
          else:
              k=1
              for i in range(2,n):
                  if s[i]==s[1]:
                      k+=1
              #print(n,k)
              #n!-C(n,k+1)*k!*(n-k+1)!
              #C(n,k+1)-c(n-1,k)=C(n-1,k+1)*(n-1)
              #print((math.factorial(n)%998244353-math.factorial(k)%998244353)%998244353)
              print((math.factorial(n)*k//(k+1))%998244353)
      
      
      