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 __future__ import division, print_function
import math
import sys
import os
from io import BytesIO, IOBase
#from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right

BUFSIZE = 8192


class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
	"""Prints the values to a stream, or to sys.stdout by default."""
	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
	at_start = True
	for x in args:
		if not at_start:
			file.write(sep)
		file.write(str(x))
		at_start = False
	file.write(kwargs.pop("end", "\n"))
	if kwargs.pop("flush", False):
		file.flush()


if sys.version_info[0] < 3:
	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

def inp():
    return(int(input()))
def inps():
    return input().strip()
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))
maxn=998244353
fac=[1,1]
for i in range(2,2*10**5+5):
    fac.append((fac[i-1]*i)%maxn)


for _ in range(inp()):
    n=inp()
    l=inlt()
    s=l.copy()
    s.sort()
    if s[-1]-s[-2]>1:
        print(0)
    else:
        mx=max(l)
        cnt=0
        for i in range(n):
            if l[i]==mx:
                cnt+=1
        if cnt>1:
            # print(n,l)
            print(fac[n])
        else:
            cnt=0
            for i in range(n):
                if mx-l[i]>=2:
                    cnt+=1
            ans=fac[n]
            for j in range(cnt+1):
                ans=(ans-(fac[n-j-1]*(fac[cnt]//fac[cnt-j]))%maxn)%maxn
                # print(fac[n-j-1]*(fac[cnt]//fac[cnt-j]))
            print(ans%maxn)