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.
from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right

import sys

def input():
    return sys.stdin.readline().rstrip()
def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getListGraph():
    return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]

mod = 10 ** 9 + 7
MOD = 998244353
# sys.setrecursionlimit(1000000)
inf = float('inf')
eps = 10 ** (-10)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]

#############
# Main Code #
#############

mod = 998244353

lim = 10 ** 6 + 1
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1]

for i in range(2, lim + 1):
    fact.append((fact[-1] * i) % mod)
    inv.append((-inv[mod % i] * (mod // i)) % mod)
    # 累計
    factinv.append((factinv[-1] * inv[-1]) % mod)

# 階乗
def factorial(n, r):
    if (r < 0) or (n < r):
        return 0
    return fact[n] * factinv[n - r] % mod

T = getN()
for _ in range(T):
    N = getN()
    A = getList()
    d = defaultdict(int)
    for a in A:
        d[a] += 1
    ma = max(A)

    if d[ma] > 1:
        print(fact[N])
    else:
        if d[ma - 1] == 0:
            print(0)
        else:
            # d[ma] - 1
            print((factorial(N, 1 + d[ma - 1]) * d[ma - 1] * pow(1 + d[ma - 1], mod - 2, mod) * fact[N - (1 + d[ma - 1])]))
