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.
from collections import defaultdict
# import math
div = 998244353
def factorial(n):
    p = 1
    for i in range(1,n+1):
        p = p*i%div
    return p

def semi_factorial(n, q):
    p = 1
    for i in range(1,n+1):
        if i == q+1:
            p=p*q%div
        else:
            p = p*i%div
    return p
def solve(n,A):
    max_a, min_a = max(A), min(A)
    d = defaultdict(int)
    for a in A:
        d[a] += 1
    if d[max_a] == 1:
        if d[max_a-1] == 0:
            return 0
        return (semi_factorial(n, d[max_a-1]))%div
    # if d[max_a] == 1:
    #     return 0
    # ans = 1
    # for k, v in d.items():
    #     ans = ans*math.factorial(v)%div
    return factorial(n)%div

def main():
    ans = []
    t = int(input())
    for _ in range(t):
        n = int(input())
        A = list(map(int, input().split(' ')))
        ans.append(solve(n,A))
    for a in ans:
        print(a)


if __name__ == '__main__':
    main()