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.
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()