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;
  int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
  long long fact[200005];
  void comp() {
    fact[0] = 1;
    for (int i = 0; i < 200005 - 1; i++) {
      fact[i + 1] = ((i + 1) * fact[i]) % 998244353;
    }
  }
  long long exp(long long n, long long x) {
    if (x == 0) return 1;
    if (x == 1) return n;
    long long tmp = exp(n, x / 2);
    tmp *= tmp;
    tmp %= 998244353;
    if (x % 2 == 1) tmp *= n;
    return tmp % 998244353;
  }
  int main() {
    ios::sync_with_stdio(0);
    int t;
    cin >> t;
    comp();
    for (int q = 0; q < t; q++) {
      int n;
      cin >> n;
      vector<int> arr;
      for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        arr.push_back(x);
      }
      sort(arr.begin(), arr.end(), greater<int>());
      if (arr[0] == arr[1] + 1) {
        int cnt = 1;
        while (cnt < n && arr[cnt] == arr[1]) cnt++;
        cnt--;
        cout << ((fact[n] * cnt) % 998244353 * exp(cnt + 1, 998244353 - 2)) %
                    998244353
             << '\n';
      } else if (arr[0] == arr[1])
        cout << fact[n] << '\n';
      else
        cout << "0\n";
    }
  }
  