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 main() {
    int t;
    cin >> t;
    for (int ii = 0; ii < t; ii++) {
      long long n;
      cin >> n;
      long long A[200000];
      for (long long i = 0; i < n; i++) {
        cin >> A[i];
      }
      long long mmax = A[0];
      for (long long i = 1; i < n; i++) {
        mmax = max(mmax, A[i]);
      }
      long long b = 0;
      for (long long i = 0; i < n; i++) {
        if (A[i] == mmax) b++;
      }
      if (b > 1) {
        long long c = 1;
        for (long long i = 2; i <= n; i++) {
          c *= i;
          c %= 998244353;
        }
        cout << c << endl;
        continue;
      }
      b = 0;
      for (long long i = 0; i < n; i++) {
        if (A[i] == mmax - 1) b++;
      }
      if (b > 0) {
        {
          n--;
          int k = n - b;
          long long N[200000], B[200000];
          long long c = 1;
          for (long long i = 1; i <= n + 1; i++) {
            c *= i;
            c %= 998244353;
            N[i] = c;
          }
          c = 1;
          for (long long i = k; i >= 1; i--) {
            c *= i;
            c %= 998244353;
            B[i] = c;
          }
          long long s = N[n];
          long long h;
          for (long long i = 0; i < k; i++) {
            h = (B[k - i] * N[n - 1 - i]) % 998244353;
            s += h;
            s %= 998244353;
          }
          long long ff = N[n + 1] - s;
          if (ff <= 0) ff += 998244353;
          cout << ff << endl;
        }
        continue;
      } else {
        cout << 0 << endl;
      }
    }
    return 0;
  }
  