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 os
import sys
from io import BytesIO, IOBase

_print = print
BUFSIZE = 8192


def dbg(*args, **kwargs):
    _print('\33[95m', end='')
    _print(*args, **kwargs)
    _print('\33[0m', end='')


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 inp():
    return sys.stdin.readline().rstrip()


def mpint():
    return map(int, inp().split(' '))


def itg():
    return int(inp())


# ############################## import
# 2020/11/1
MAX_N = 3 * 10 ** 5
MOD = 998244353

MAX_N = min(MAX_N, MOD - 1)
fac = [1] * (MAX_N + 1)
for i in range(MAX_N):
    fac[i + 1] = fac[i] * (i + 1) % MOD

A000142 = fac.__getitem__
# ############################## main
from heapq import nlargest


def solve():
    n = itg()
    arr = tuple(mpint())
    b, a = nlargest(2, arr)
    if a == b:
        return A000142(n)
    if b - a != 1:
        return 0
    m = arr.count(a)
    # a, a, ..., b is not nice
    # there are 1/(m+1) are not nice
    # only m/(m+1) are nice
    return A000142(n) * m % MOD * pow(m + 1, MOD - 2, MOD) % MOD


def main():
    # print(solve())
    for _ in range(itg()):
        print(solve())
        # solve()
        # print("YES" if solve() else "NO")
        # print("yes" if solve() else "no")


DEBUG = 0
URL = 'https://codeforces.com/contest/1569/problem/C'

if __name__ == '__main__':
    # 0: normal, 1: runner, 2: debug, 3: interactive
    if DEBUG == 1:
        import requests
        from ACgenerator.Y_Test_Case_Runner import TestCaseRunner

        runner = TestCaseRunner(main, URL)
        inp = runner.input_stream
        print = runner.output_stream
        runner.checking()
    else:
        if DEBUG != 2:
            dbg = lambda *args, **kwargs: ...
            sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
        if DEBUG == 3:
            def print(*args, **kwargs):
                _print(*args, **kwargs)
                sys.stdout.flush()
        main()
# Please check!
