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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class JuryMeeting {
      static long[] factorial;
      static long[] invfact;
      static int MOD=998244353;
      static int N=(int)(2*1e5)+5;
  
      public static long nPr(int n, int r) {
          return (n < r) ? 0 : (factorial[n] * invfact[n - r]) % MOD;
      }
      public static void precompute()
      {
          factorial[0]=factorial[1]=1;
          for(int i=2;i<N;i++)
          {
              factorial[i]=factorial[i-1]*i;
              factorial[i]%=MOD;
          }
          invfact[N-1]=modinv(factorial[N-1]);
          for(int i=N-2;i>=0;i--)
          {
              invfact[i]=invfact[i+1]*(i+1);
              invfact[i]%=MOD;
          }
      }
      static long modinv(long k)
      {
          return pow(k, MOD-2, MOD);
      }
      static long pow(long a, long b, long m)
      {
          long ans=1;
          while(b>0)
          {
              if(b%2!=0)
                  ans=(ans*a)%m;
              b/=2;
              a=(a*a)%m;
          }
          return ans;
      }
  
      public static void main(String args[]) {
          FastScanner sc = new FastScanner();
          int test = sc.nextInt();
          factorial=new long[200000+5];
          invfact=new long[200000+5];
          precompute();
  
          while (test-- != 0) {
              int n=sc.nextInt();
              ArrayList<Integer> arr=new ArrayList<>(n);
  
              for(int i=0;i<n;++i){
                  arr.add(sc.nextInt());
              }
              Collections.sort(arr);
  
              if(arr.get(n-1).equals(arr.get(n-2))){
                  System.out.println(factorial[n]);
                  continue;
              }
              if(arr.get(n-1)-arr.get(n-2)>=2) {
                  System.out.println(0);
                  continue;
              }
  
              int numberOfSecondHighest=1;
  
              for(int i=n-3;i>=0;--i){
                  if(arr.get(i).equals(arr.get(n-2)))
                      numberOfSecondHighest++;
                  else
                      break;
              }
              // np0, np1, np2
              // 1,  n, n(n-1)
              long answer=0;
  //            long permute=n-numberOfSecondHighest-1;
  //            long permutation=permute;
              for(int i=1;i<=n-numberOfSecondHighest;++i){
  
                  answer=(answer+(factorial[n-i]*nPr(n-numberOfSecondHighest-1,i-1))%998244353)%998244353;
  //                permute--;
  //                permutation=((permutation%998244353)*(permute%998244353))%998244353;
              }
  
              System.out.println(((factorial[n]-answer)+998244353)%998244353);
          }
      }
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  