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 gcd(a, b):
    if a < b:
        b, a = a, b
    while a * b > 0:
        a %= b
        b, a = a, b
    return a


def candy(k, c):
    if gcd(k, c) > 1:
        return -1
    s = [1, 0]
    r = [c, k]
    q = [0]
    i = 2
    while r[i - 1] != 0:
        q.append(r[i - 2] // r[i - 1])
        s.append(s[i - 2] - s[i - 1] * q[i - 1])
        r.append(r[i - 2] - r[i - 1] * q[i - 1])
        i += 1
    return s[i - 2] % k


def sex(n, a):
    m = max(a)
    idx = a.index(m)
    okay = False
    for i in range(n):
        if i != idx and m - a[i] <= 1:
            okay = True
            break
    if not okay:
        return 0
    okay = False
    for i in range(idx + 1, n):
        if a[i] == m:
            okay = True
            break
    if okay:
        return fac[n]
    tot = 0
    for i in range(n):
        if a[i] == m - 1:
            tot += 1
    return (fac[n] - fac[n] * candy(p, tot + 1)) % p


p = 998244353
fac = [1] * 200001
for i in range(2, 200001):
    fac[i] = fac[i - 1] * i
    fac[i] %= p
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    print(sex(n, a))