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
  
  #comment these out later
  #sys.stdin = open("in.in", "r")
  #sys.stdout = open("out.out", "w")
  
  input = sys.stdin.readline
  
  def main():
  
  	def extended_gcd(a, b):
  	    """returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
  	    s, old_s = 0, 1
  	    r, old_r = b, a
  	    while r:
  	        q = old_r // r
  	        old_r, r = r, old_r - q * r
  	        old_s, s = s, old_s - q * s
  	    return old_r, old_s, (old_r - old_s * a) // b if b else 0
  
  
  	def modinv(a, m):
  	    """returns the modular inverse of a w.r.t. to m, works when a and m are coprime"""
  	    g, x, _ = extended_gcd(a % m, m)
  	    return x % m if g == 1 else None
  
  	MOD = 998244353 
  
  	fact = [1]
  
  	for x in range(1, 200005):
  		fact.append((fact[-1] * x)%MOD)
  
  	def comb(n, r):
  		modi = modinv(fact[r] * fact[n-r], MOD)		
  		return (fact[n] * modi)%MOD
  
  	#print(comb(1000, 1))
  
  	t = int(input())
  	#t = 1
  
  	for _ in range(t):
  		n = int(input())
  		ar = list(map(int, input().split()))
  
  		m = max(ar)
  
  		if ar.count(m) > 1:
  			print(fact[n])
  
  		else:
  			m2c = ar.count(m-1)
  
  			ans = 0
  
  			for x in range(m2c, n):
  				#print(sar, x, n-x-1)
  				toadd = (fact[x] * fact[n - x - 1])%MOD
  
  				#print(toadd)
  
  				toadd *= comb(n - m2c - 1, x - m2c)
  				#print(x, toadd)
  
  				#print(n-x, x-m2c, toadd)
  
  				ans += toadd%MOD
  
  				ans%= MOD
  
  			print((fact[n] - ans)%MOD)
  
  
  main()