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.path
from collections import *
import math
import bisect
import heapq as hq
from fractions import Fraction
from random import randint

if (os.path.exists('input.txt')):
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")


##########################################################

input = sys.stdin.readline
mod = 998244353

fact = [1] * (100000)

for i in range(1,len(fact)):
    fact[i] = (fact[i - 1] * i) % mod

def power(x, y, m):
 
    if (y == 0):
        return 1
 
    p = power(x, y // 2, m) % m
    p = (p * p) % m
 
    if(y % 2 == 0):
        return p
    else:
        return ((x * p) % m)

def modInverse(a):
    return power(a, mod - 2, mod)

def solve():
    n = int(input())
    arr = [int(x) for x in input().split()]
    arr.sort()
    if arr[-1] - arr[-2] > 1:
        print(0)
        return
    if arr[-1] == arr[-2]:
        print(fact[n])
        return

    res = 0
    x = arr.index(arr[-2])

    num = 1
    den = 1
    res = 0

    for i in range(x):
        num = (num * (x - i)) % mod
        den = (den * (i + 1)) % mod
        y = 1
        y = (y * fact[n - (i + 2)]) % mod
        y = (y * fact[i + 1]) % mod
        y = (y * num) % mod
        y = (y * modInverse(den)) % mod
        res = (res + y) % mod
    print(fact[n] - fact[n - 1] - res)

t = int(input())
while t:
    t -= 1
    solve()

##########################################################
