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.
import sys
import os
from io import BytesIO

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))

f = sys.stdin
if os.environ.get('USER') == "loic":
    f = open("data.in")

line = lambda: f.readline().strip('\r\n').split()

def write(w):
    sys.stdout.write(w)
    sys.stdout.write("\n")

def factorial_modulo(limit):
    fact = [1]
    for i in range(1, limit+1):
        fact.append(fact[-1] * i % M)
    return fact

def inv_modulo_all(limit):
    mod_inv = [0,1]
    for i in range(2, limit+1):
        mod_inv.append(M - (M // i) * mod_inv[M % i] % M)
    return mod_inv

def factorial_modulo_inverse(limit):
    mod_inv = inv_modulo_all(limit)
    fact_inv = [1,1]
    for i in range(2, limit+1):
        fact_inv.append(fact_inv[-1] * mod_inv[i] % M)
    return fact_inv

def solve():
    d = sorted(D,reverse=True)
    a = d[0]
    b = d[1]
    c = 1
    for i in range(2,N):
        if d[i] == b:
            c += 1
        else:
            break
    
    if a - b > 1:
        return str(0)
    if a==b:
        return str(FACT[N])
    
    res = c * FACT[N-1]
    res %= M
    
    for i in range(c,N-1):
        tmp = FACT[i] * FACT_INV[i-c] * FACT[N-1-c]
        res += FACT[N-1] - tmp
        res %= M
    
    return str(res)

M = 998244353
Z = 200000
FACT = factorial_modulo(Z)
FACT_INV = factorial_modulo_inverse(Z)

T = int(line()[0])
for test in range(1,T+1):
    N = int(line()[0])
    D = list(map(int,line()))
    write(solve())
    
f.close()