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: import java.util.Scanner;
  
  public class C {
      static int MOD = 998244353;
  
      public static void main(String[] args) {
          Scanner scn = new Scanner(System.in);
          long[] fac = new long[(int) 2e5 + 2];
          long[] invFac = new long[(int) 2e5 + 2];
          int test = scn.nextInt();
          fac[0] = 1;
          invFac[0] = binpow(fac[0], MOD - 2, MOD);
          for (int i = 1; i < fac.length; i++) {
              fac[i] = 1l * fac[i - 1] * i;
              fac[i] %= MOD;
              invFac[i] = binpow(fac[i], MOD - 2, MOD);
          }
          while (test-- > 0) {
              int n = scn.nextInt();
              int[] arr = new int[n];
              int sndMaxCount = 0;
              int maxCount = 0;
  
              int max = -1;
              int secondmax = -1;
              for (int i = 0; i < n; i++) {
                  arr[i] = scn.nextInt();
                  if (arr[i] == secondmax) sndMaxCount++;
                  if (arr[i] == max) maxCount++;
                  if (arr[i] > max) {
                      secondmax = max;
                      max = arr[i];
                      sndMaxCount = maxCount;
                      maxCount = 1;
                  }
                  if (arr[i] < max && arr[i] > secondmax) {
                      sndMaxCount = 1;
                      secondmax = arr[i];
                  }
  
              }
  
              if (maxCount > 1) {
                  System.out.println(fac[n]);
                  continue;
              }
              if (secondmax < max - 1) {
                  System.out.println(0);
                  continue;
              }
              long ans = 0;
              for (int i = 0; i < n - 1; i++) {
                  if (i < sndMaxCount) {
                      ans += fac[n - 1];
                      ans %= MOD;
  
                  } else {
                      long add = fac[n - 1];
                      long c = 1l * fac[(n - 1 - sndMaxCount)] * invFac[(n - 1 - sndMaxCount) - (n - 1 - i)];
                      c %= MOD;
                      c *= fac[i];
                      c %= MOD;
                      ans += (add - c) % MOD;
                      ans %= MOD;
                      if (ans < 0) ans += MOD;
  
                  }
              }
              System.out.println(ans % MOD);
  
          }
  
      }
  
      public static long binpow(long x, long n, long m) {
          assert (n >= 0);
          x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow
          long res = 1;
          while (n > 0) {
              if (n % 2 == 1)  // if n is odd
                  res = res * x % m;
              x = x * x % m;
              n /= 2;  // divide by two
          }
          return res;
      }
  
  }