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.*;
  
  public class c {
  
      static long fact[] = new long[200005];
      static long M = (long)998244353;
  
      public static long bin_exp(long a, long b){
  
          if(b == 0){
              return 1;
          }
  
          long x = bin_exp(a, b/2);
  
          if(b%2 == 0){
              long ans = (x*x)%M;
              return ans;
          }
  
          else{
              long ans = a*(x*x%M)%M;
              return ans;
          }
  
      }
  
      public static long inv(long a){
  
          return bin_exp(a, M-2);
  
      }
  
      public static long perm(int a, int b){
  
          return (fact[a] * inv(fact[a-b]))%M;
  
      }
  
      public static long binom(int a, int b){
  
          if (b < 0 || a < 0)return 0;
          return (fact[a] * ((inv(fact[b]) * inv(fact[a-b]))%M)%M)%M;
  
      }
  
      public static void main(String[] args){
  
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
  
          fact[0] = 1L;
          for(int i=1; i<fact.length; i++){
              fact[i] = (fact[i-1]*i)%M;
          }
  
  
          while(t-- > 0){
  
              int n = sc.nextInt();
              int arr[] = new int[n];
  
              for(int i=0; i<n; i++){
                  arr[i]  =sc.nextInt();
              }
  
              int maxVal = arr[0];
              int secondMaxVal = -1;
              for(int i=1; i<n; i++){
                  if(arr[i] > maxVal){
                      secondMaxVal = maxVal;
                      maxVal = arr[i];
                  }
                  else if(arr[i] > secondMaxVal){
                      secondMaxVal = arr[i];
                  }
              }
  
              if(maxVal == secondMaxVal){
                  System.out.println(fact[n]);
                  continue;
              }
  
              if(maxVal - secondMaxVal > 1){
                  System.out.println(0);
                  continue;
              }
  
              int count = 0;
              for(int i=0; i<n; i++){
  
                  if(arr[i] == secondMaxVal){
                      count++;
                  }
  
              }
  
  
              long ans = fact[n];
              ans = (ans - fact[n-1]);
              if(ans < 0){
                  ans += M;
              }
              ans = ans%M;
              int iter = n-count-1;
  
              long sum = 0L;
              for(int i=1; i<=iter; i++){
                  sum = (sum + (perm(iter, i) * fact[n-1-i])%M)%M;
              }
  
              ans  = ans - sum;
              if(ans < 0){
                  ans += M;
              }
              ans = ans%M;
  
              System.out.println(ans);
  
          }
  
      }
  
  }