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 itertools
from sys import maxsize, stdout, stdin,stderr
mod = int(1e9+7)
import sys
def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return list(map(str,input().strip()))
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter, deque
import math
import heapq
from heapq import heappop , heappush
import bisect
from math import factorial, inf
from itertools import groupby
from itertools import permutations as comb
def gcd(a,b): 
    while b:
        a %= b
        tmp = a
        a = b
        b = tmp
    
    return a
 
def lcm(a,b): 
    return a  // gcd(a, b) * b
 
def check_prime(n):
    for i in range(2, int(n ** (1 / 2)) + 1):
        if not n % i:
            return False
    return True

def nCr(n, r):
 
    return (fact(n) // (fact(r)
                * fact(n - r)))
 
# Returns factorial of n
def fact(n):
 
    res = 1
     
    for i in range(2, n+1):
        res = res * i
         
    return res
def primefactors(n):
    num=0
    
    while n % 2 == 0:
        num+=1
        n = n / 2
    
    for i in range(3,int(math.sqrt(n))+1,2):
         
    
        while n % i== 0:
            num+=1
            n = n // i
               
    
    if n > 2:
        num+=1
    return num
'''
def iter_ds(src):
    store=[src]
    while len(store):
        tmp=store.pop()
        if not vis[tmp]:
            vis[tmp]=True
            for j in ar[tmp]:
                store.append(j)
'''
def ask(a,b,c):
    # print('? 1 {}'.format(a),flush=True)
    print(c,a,b,flush=True)
    n=I()
    
    return n
def linear_sieve(n):
    is_composite=[False]*n
    prime=[]
    for i in range(2,n):
        if not is_composite[i]:
            prime.append(i)
        for j in prime:
            is_composite[i*j]=True
            if i%prime==0:
                break
    return prime
 
def dfs(i,p,d):
    
    a,tmp=0,0
    for j in d[i]:
        if j!=p:
            a+=1
            tmp+=dfs(j,i)
    
    if a==0:
        return 0
    
    return tmp/a + 1 
def primeFactors(n):
    l=[]
    
    while n % 2 == 0:
        l.append(2)
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2):
         
        
        while n % i== 0:
            l.append(i)
            n = n // i

    if n > 2:
        l.append(n)
    return l


    

#  Sieve
d=[]
primes=[]
prim=[0]*(10**5+1)
def sieve(n):
    
    for i in range(n):
        d.append(i)

    for i in range(2,n):

        if d[i]==i:
            prim[i]=1
            primes.append(i)
        j=0
        while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
        
            d[i * primes[j]] = primes[j]
            j+=1

        
def primeFactors(n):
    factors=[]
    while n!=1:
        factors.append(d[n])
        n//=d[n]
    
    return factors
def func(a):
    b=1
    res=0
    for i in range(len(a)-1,-1,-1):
        res+=b*int(a[i])
        b*=2
    return res

 

t = 1 
t = I()
for _ in range(t):
    n = I()
    s= lint()
    s.sort()
    d=Counter(s)
    if s[n-1]-s[n-2]>=2:
        print(0)
    else:
        if s[n-1]!=s[n-2]:
            tmp=d[s[n-2]]
            
            if n>=mod:
                print(0)
            else:
                res=1
                for i in range(1,n+1):
                    if i==tmp+1:
                        continue
                    if i==tmp:
                        res=(res*i)%mod
                    res=(res*i)%mod
                print(res)

        else:
        
            print(factorial(n))
        