description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mxN = 2e5 + 7;
  const long long MOD = 998244353;
  int a[mxN];
  long long silnia[mxN];
  long long potega(long long x) {
    long long res = 1;
    long long y = MOD - 2;
    while (y) {
      if (y & 1) res = (res * x) % MOD;
      x = (x * x) % MOD;
      y /= 2;
    }
    return res;
  }
  void solve() {
    int n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    sort(a + 1, a + n + 1);
    int ile = 0;
    for (long long i = 1; i <= n; i++)
      if (a[i] == a[n - 1]) ile++;
    if (a[n] == a[n - 1])
      cout << silnia[n] << "\n";
    else if (a[n] - a[n - 1] == 1) {
      long long x = (silnia[n] * potega(ile + 1)) % MOD;
      cout << (silnia[n] - x + MOD) % MOD << "\n";
    } else
      cout << "0\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    silnia[0] = 1;
    for (long long i = 1; i <= 2e5; i++) silnia[i] = (silnia[i - 1] * i) % MOD;
    int t;
    cin >> t;
    while (t--) solve();
  }
  