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 itertools
  from sys import maxsize, stdout, stdin,stderr
  mod = int(1e9+7)
  import sys
  def I(): return int(stdin.readline())
  def lint(): return [int(x) for x in stdin.readline().split()]
  def S(): return list(map(str,input().strip()))
  def grid(r, c): return [lint() for i in range(r)]
  from collections import defaultdict, Counter, deque
  import math
  import heapq
  from heapq import heappop , heappush
  import bisect
  from math import factorial, inf
  from itertools import groupby
  from itertools import permutations as comb
  def gcd(a,b): 
      while b:
          a %= b
          tmp = a
          a = b
          b = tmp
      
      return a
   
  def lcm(a,b): 
      return a  // gcd(a, b) * b
   
  def check_prime(n):
      for i in range(2, int(n ** (1 / 2)) + 1):
          if not n % i:
              return False
      return True
  
  def nCr(n, r):
   
      return (fact(n) // (fact(r)
                  * fact(n - r)))
   
  # Returns factorial of n
  def fact(n):
   
      res = 1
       
      for i in range(2, n+1):
          res = res * i
           
      return res
  def primefactors(n):
      num=0
      
      while n % 2 == 0:
          num+=1
          n = n / 2
      
      for i in range(3,int(math.sqrt(n))+1,2):
           
      
          while n % i== 0:
              num+=1
              n = n // i
                 
      
      if n > 2:
          num+=1
      return num
  '''
  def iter_ds(src):
      store=[src]
      while len(store):
          tmp=store.pop()
          if not vis[tmp]:
              vis[tmp]=True
              for j in ar[tmp]:
                  store.append(j)
  '''
  def ask(a,b,c):
      # print('? 1 {}'.format(a),flush=True)
      print(c,a,b,flush=True)
      n=I()
      
      return n
  def linear_sieve(n):
      is_composite=[False]*n
      prime=[]
      for i in range(2,n):
          if not is_composite[i]:
              prime.append(i)
          for j in prime:
              is_composite[i*j]=True
              if i%prime==0:
                  break
      return prime
   
  def dfs(i,p,d):
      
      a,tmp=0,0
      for j in d[i]:
          if j!=p:
              a+=1
              tmp+=dfs(j,i)
      
      if a==0:
          return 0
      
      return tmp/a + 1 
  def primeFactors(n):
      l=[]
      
      while n % 2 == 0:
          l.append(2)
          n = n // 2
      for i in range(3,int(math.sqrt(n))+1,2):
           
          
          while n % i== 0:
              l.append(i)
              n = n // i
  
      if n > 2:
          l.append(n)
      return l
  
  
      
  
  #  Sieve
  d=[]
  primes=[]
  prim=[0]*(10**5+1)
  def sieve(n):
      
      for i in range(n):
          d.append(i)
  
      for i in range(2,n):
  
          if d[i]==i:
              prim[i]=1
              primes.append(i)
          j=0
          while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
          
              d[i * primes[j]] = primes[j]
              j+=1
  
          
  def primeFactors(n):
      factors=[]
      while n!=1:
          factors.append(d[n])
          n//=d[n]
      
      return factors
  def func(a):
      b=1
      res=0
      for i in range(len(a)-1,-1,-1):
          res+=b*int(a[i])
          b*=2
      return res
  
   
  mod=998244353
  t = 1 
  t = I()
  for _ in range(t):
      n = I()
      s= lint()
      s.sort()
      d=Counter(s)
      if s[n-1]-s[n-2]>1:
          print(0)
      elif s[n-1]!=s[n-2]:
              tmp=d[s[n-2]]
              
              if n>=mod:
                  print(0)
              else:
                  res=1
                  for i in range(1,n+1):
                      if i==tmp+1:
                          continue
                      if i==tmp:
                          res=(res*i)%mod
                      res=(res*i)%mod
                  print(res%mod)
  
      else:
          
          print(factorial(n)%mod)
          