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 a[1111111];
  int q;
  int n;
  int64_t gt[1111111];
  int64_t Powe(int a, int n) {
    if (n == 0) return 1;
    if (n == 1) return a;
    int64_t b = Powe(a, n / 2);
    b = (b * b) % 998244353;
    if (n % 2 == 1)
      return (b * a) % 998244353;
    else
      return b;
  }
  int main() {
    scanf("%d", &q);
    gt[0] = 1;
    for (int i = 1; i <= 1000000; i++) gt[i] = (gt[i - 1] * i) % 998244353;
    for (int z = 1; z <= q; z++) {
      scanf("%d", &n);
      int an, as = 0;
      bool ok = true;
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (as < a[i]) {
          an = as;
          as = a[i];
        } else
          an = max(an, a[i]);
      }
      sort(a + 1, a + n + 1);
      int A = upper_bound(a + 1, a + n + 1, a[n]) -
              lower_bound(a + 1, a + n + 1, a[n]);
      if (A > 1) {
        cout << gt[n] << endl;
        ok = false;
      }
      if (a[n] - an > 1) {
        cout << 0 << endl;
        ok = false;
      }
      if (ok) {
        int B =
            upper_bound(a + 1, a + n + 1, an) - lower_bound(a + 1, a + n + 1, an);
        int64_t dd = (gt[B] * gt[n - B - 1]) % 998244353;
        int T = n - B - 1;
        int64_t X = 1;
        for (int i = B + 2; i <= n; i++) {
          int u = i - B - 1;
          int64_t cc = (gt[u] * gt[T - u]) % 998244353;
          cc = (gt[T] * Powe(cc, 998244353 - 2)) % 998244353;
          dd = (dd + (((gt[i - 1] * cc) % 998244353) * gt[n - i]) % 998244353) %
               998244353;
        }
        cout << (gt[n] - dd + 998244353) % 998244353 << endl;
      }
    }
  }
  