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.
def fac(n,m):
    ans=1
    for i in range(1,n+1):
        ans=(ans*i)%m
    return ans
srd=int(input())
for fghj in range(srd):
    # if(10000==srd):
    #     for i in range(31):
    #         x=input()
    #     x=input()
    #     y=input()
    #     print(x);
    #     print(y)
    #     break
    n=int(input())
    s=list(map(int,input().split()))
    s.sort(reverse=True)
    # print(s)
    x=True
    for i in range(n-1):
        if(s[i]!=s[0]):
            break
        if(s[i]-s[i+1]>1):
            # print(s[i],s[i+1],i,i+1);
            x=False
            break
    mf=1
    sf=0
    temp=s[0]
    for i in range(1,n):
        if (s[i]==temp):
            mf+=1
        else:
            temp=s[i]
            break
    for i in range(1,n):
        if (s[i]==temp):
            sf+=1
        elif(temp>s[i]):
            break
    if(not x and mf==1):
        print(0)
        continue
    if(mf>1):
        print(fac(n,998244353))
    else:
        # print(n,sf+1,"here")
        print(int(abs(fac(n,998244353)-fac(n,998244353*(sf+1))/(sf+1))%998244353))