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
input = sys.stdin.readline

MOD = 998244353

facto = dict()
def initFacto(N):
	facto[0] = 1
	facto[1] = 1
	for n in range(2, N+1):
		facto[n] = (n * facto[n-1]) % MOD

def solve():
	N = int(input())
	nums = list(map(int, input().split()))

	uniq = list(set(nums))
	uniq.sort()

	first = uniq[-1]
	firstCnt = nums.count(first)

	all_cases = facto[N]
	if len(uniq) == 1 or firstCnt >= 2:
		print(all_cases)
		return

	second = uniq[-2]
	secondCnt = nums.count(second)

	if first-second != 1:
		print(0)
		return

	rest = N - 1 - secondCnt
	# 1 2 3 3 3 4 4

	insertPt = secondCnt + 2

	# 순열 조합!
	check = facto[insertPt-1 + rest]//facto[insertPt-1]
	ans = (facto[secondCnt] * (facto[insertPt-1 + rest]//facto[insertPt-1])) % MOD
	print((all_cases - ans + MOD) % MOD)


if __name__ == '__main__':
	T = int(input())
	initFacto(int(2*1e5))
	for _ in range(T):
		solve()