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.
def solve(n,tasks):
    mod_val = 998244353
    max1 = 0
    max2 = 0
    for t in tasks:
        if t > max1:
            max2 = max1
            max1 = t
        elif t > max2:
            max2 = t
    if max1 > max2 + 1:
        return 0
    if max1 == max2+1:
        count2 = tasks.count(max2)
        tot = get_factorial(n, mod_val)
        for i in range(n):
            if i >= count2:
                tot-=choose(n-count2-1,i-count2, mod_val) *get_factorial(n-i-1, mod_val)*get_factorial(i, mod_val)
                # print(tot)
        return tot
    res = get_factorial(n, mod_val)

    # print(tasks)
    return res

fac_cache={}
inv_fac_cache = {}
mod_val = 998244353

def invert_mod(i, mod_val):
    return pow(i, mod_val-2, mod_val)

def choose(N,k, mod_val):
    if k == 0:
        return 1
    if k == N:
        return 1
    if k == 1:
        return N
    if k == (N-1):
        return N
    n_fac = get_factorial(N, mod_val)
    nk_inv_fac = get_inv_factorial(N-k, mod_val)
    k_inv_fac = get_inv_factorial(k, mod_val)
    res = (n_fac * nk_inv_fac) % mod_val
    res = (res * k_inv_fac) % mod_val
    return res

# if you run the biggest factorial first then all are cached
def get_factorial(N, mod_val):
    cached_res = fac_cache.get(N, None)
    if cached_res:
        return cached_res
    res = 1
    for i in range(2, N+1):
        res = (res * i) %mod_val
        fac_cache[i] = res
    return res


def get_inv_factorial(k, mod_val):
    cached_res = inv_fac_cache.get(k, None)
    if cached_res:
        return cached_res
    res = 1
    for i in range(2, k+1):
        res = (res * invert_mod(i, mod_val)) % mod_val
        inv_fac_cache[i] = res
    return res

import os
import io
import time
if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    T = int(input().decode().strip())
    for t in range(T):
        n = int(input().decode().strip())
        tasks = [int(x) for x in input().decode().strip().split(" ")]
        res = solve(n,tasks)
        print(res)