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.


mod=998244353
 
def make_mod_mul(mod = 10**9 + 7):
    fmod_inv = 1.0 / mod
    def mod_mul(a, b, c=0):
        res = a * b +  c - mod *  int(fmod_inv * a * b + fmod_inv * c)
        if res >= mod:
            return res - mod
        elif res < 0:
            return res + mod
        else:
            return res
    return mod_mul
 
mod_mul=make_mod_mul(mod)
def mod_pow(x, y):
    if y == 0:
        return 1
    res = 1
    while y > 1:
        if y & 1 == 1:
            res = mod_mul(res,x)
        x = mod_mul(x, x)
        y >>= 1
    return mod_mul(res, x)
 
MAX=200005
fact=[1]
for zz in range(1,MAX+1):
    fact.append(mod_mul(fact[-1],zz))
def nCr(n,r): #choose, MOD is prime
    num=fact[n]
    den=mod_mul(fact[r],fact[n-r])
    return (num*pow(den,mod-2, mod))%mod
 
    
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    a = sorted(a)
    ans = 'yes'
    m = max(a)
    mm = a[0]
    for i in range(1, len(a)):
        if a[i] > mm and a[i] != m:
            mm = a[i]

    if m-mm>1:
        print(0)
    elif m == mm:
        print(fact[n])
    elif a[-1] == a[-2] and n == 2:
        print(2)
    elif a[-1] == a[-2]:
        print(fact[n])
    else:
        k = 0
        
        for i in range(len(a)):
            if a[i] == mm:
                k += 1
          
        ans = fact[n]
        for i in range(k, n):
            #print(i, ans,nCr(i, i-k) * fact[k])
            ans = (ans + mod - nCr(i, i-k) * fact[k]*fact[n-k-1])%mod
            
        if n == 2:
            print(1)
        else:
            print(ans)