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 heapq,math
from collections import defaultdict,deque
from os import getcwd
from itertools import permutations
#from functools import cmp_to_key

import sys, os.path
#sys.setrecursionlimit(10000000)

if(os.path.exists('C:/Users/Dhanush/Desktop/cp/input.txt')):
    sys.stdout = open('C:/Users/Dhanush/Desktop/cp/output.txt', 'w')
    sys.stdin = open('C:/Users/Dhanush/Desktop/cp/input.txt', 'r')

input=sys.stdin.readline
mod=998244353
def isnice(l):
    i=0
    prev=-1
    cnt=n
    while(cnt>0):
        if(l[i]==1):
            if(prev==i):
                return False
            l[i]=0
            cnt-=1
            prev=i
        elif(l[i]>1):
            if(prev==i):
                return False
            l[i]-=1
            prev=i
        else:
            if(prev==i):
                return True
        i=(i+1)%n


    return True

factarr=[1]
maxi=2*(10**5)

res=1
for i in range(1,maxi+5):
    res=(res*i)%mod
    factarr.append(res)


def ncr(n,r):
    num=factarr[n]
    denom=(factarr[r]*factarr[n-r])%mod
    return num//denom


tc=int(input())
for _ in range(tc):
    n=int(input())
    l=sorted(list(map(int,input().split())))
    d=defaultdict(int)
    for i in l:
        d[i]+=1
        '''
    perm=list(permutations(l))
    print(len(perm),perm)
    c=0
    for row in perm:
        if(isnice(list(row))):
            c+=1
            print(row)
        else:
            print('not true',row)
    print(c)


        '''

    
    if(l[-1]==l[-2]):
        #ans is n!
        print(factarr[n])
    elif(l[-1]>l[-2]+1):
        print(0)
    else:
        prev=l[-2]
        count=d[prev]
        ans=0
        for i in range(count+1,n+1):
            c1=factarr[i-1]
            c2=factarr[n-i]
            tot=(c1*c2)%mod
            tot=(tot*ncr(n-count-1,i-count-1))%mod
            ans=(ans+tot)%mod
        print(factarr[n]-ans)

    