human
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;...
assistant
#Here is Python program to print it modulo 998 244 353.

import os, sys, math
from io import BytesIO, IOBase
ip = lambda: input()
it = lambda: int(input())
ma = lambda: map(int, input().split(" "))
li = lambda: list(map(int,input().split(" ")))
mod=10**9+7
md=998244353

def ncr(n):
    arr=[1]*(n+1)
    for i in range(1,n+1):
        arr[i]=arr[i-1]*(n+1-i)//i
        arr[i]%=md
    return arr

def main():
    for _ in range(int(input())):
        n=it()
        a=li()
        d=dict()
        m=1
        x = 1
        for i in range(2, n + 1):
            x = (x*i)%md
        for i in range(n):
            if a[i]>m:
                m=a[i]
            try:
                d[a[i]]+=1
            except KeyError:
                d[a[i]]=1
        if d[m]>1:
            print(x)
            continue
        if m-1 not in d:
            print(0)
            continue
        y = 1
        for i in range(2, n + 1):
            if i==d[m-1]+1:
                continue
            y = (y*i)%md
        ans=x-y
        print(ans%md)
    # Fast IO Region
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:
  ...