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.
# Bad arrangements are those where the max-1 nos are before max
import sys
input = sys.stdin.buffer.readline

MOD = 998244353
def solve():
    A.sort()
    if A[N - 1] == A[N - 2]: # (1,1) -> (0,1) -> (0,0)
        return factorial[N]

    if A[N - 1] - A[N - 2] > 1: # (2,0) -> (1,0) -> X
        return 0
    
    k = 0 # count of max - 1 elements
    for a in A:
        if a == A[N - 2]:
            k += 1

    # N-k-1 elements can be arranged in any order in the N places: (N)p(N-k-1)
    # In the remaining k+1 elements if max is at the last: k!
    bad = factorial[k] * factorial[N] % MOD * pow(factorial[k + 1], MOD - 2, MOD) % MOD
    return factorial[N] - bad


def compute_factorials(N=int(1e6)):
    fact = [1] * (N + 1)
    for i in range(2, N + 1):
        fact[i] = fact[i - 1] * i % MOD
    return fact

factorial = compute_factorials()
test_cases = int(input())
for test_case in range(test_cases):
    N = int(input())
    A = list(map(int, input().split()))
    print(solve())