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;
  long long int fac[200005];
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fac[0] = 1;
    for (long long int i = 1; i < 200005; i++)
      fac[i] = (fac[i - 1] * i) % 998244353;
    long long int t;
    cin >> t;
    for (long long int tt = 1; tt <= t; tt++) {
      long long int n;
      cin >> n;
      long long int a[n], b[n];
      for (long long int i = 0; i < n; i++) {
        cin >> a[i];
        b[i] = a[i];
      }
      sort(b, b + n);
      if (b[n - 1] - b[n - 2] > 1)
        cout << 0;
      else {
        if (b[n - 1] - b[n - 2] == 0)
          cout << fac[n];
        else {
          long long int count = 0;
          for (long long int i = n - 2; i >= 0; i--) {
            if (b[i] != b[n - 2]) break;
            count++;
          }
          long long int ans = 0;
          ans = (count % 998244353 * fac[n - 1] % 998244353) % 998244353;
          long long int x = n - 2, y = count, z = n - count;
          for (long long int i = count + 1; i < n; i++) {
            ans = (ans + ((n - i) % 998244353 *
                          (y % 998244353 * fac[x] % 998244353) % 998244353) %
                             998244353) %
                  998244353;
            x--;
            z--;
            y = (y % 998244353 * z % 998244353) % 998244353;
          }
          cout << ans;
        }
      }
      cout << endl;
    }
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  }
  