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 sys import stdin
input = stdin.readline

mod = 998244353
def mul(a , b):return ((a%mod) * (b%mod))%mod
def ncr(n , r):return mul(fact[n] , mul(pow(fact[n - r],mod - 2 , mod) , pow(fact[r] , mod - 2 , mod)))

size = (2 * (10**5)) + 1
fact = [1]
for i in range(1 , size):
    fact.append(mul(fact[-1] , i))

def answer():

    m1 , m2 = a[-1] , a[-2]

    if(m1 == m2):return fact[n]

    x1 , x2 = a.count(a[-1]) , a.count(a[-2])

    if((m1 - m2) > 1):return 0

    x = x1 + x2
    ans = mul(ncr(n , x) , mul(x1 , x2))
    ans = mul(ans , fact[x - 2])
    ans = mul(ans , x - 1)
    
    return mul(ans , fact[n - x])
    

for T in range(int(input())):

    n = int(input())

    a = sorted(list(map(int,input().split())))

    print(answer())
