human
PYTHON3 solution for "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;...":
assistant
#Here is Python program to print it modulo 998 244 353.
 def jc(n):
	t = 1
	for i in range(1, n + 1):t *= i;t %= 998244353
	return t
for _ in range(int(input())):
	n = int(input());l = [int(i) for i in input().split()];imax = max(l)
	if l.count(imax) != 1:print(jc(n))
	elif l.count(imax-1) == 0:print(0)
	else:
		x = l.count(imax - 1);y = n - 1 - x;j = jc(x);tmp = x + 2
		for i in range(y):j *= (tmp + i);j %= 998244353
		print((jc(n) - j) % 998244353)