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.*;
  import java.util.*;
  public class Main {
      public static void main(String args[])
      {
          FastReader input=new FastReader();
          PrintWriter out=new PrintWriter(System.out);
          int T=input.nextInt();
          while(T-->0)
          {
              int n=input.nextInt();
              int a[]=new int[n];
              for(int i=0;i<n;i++)
              {
                  a[i]=input.nextInt();
              }
              long m=998244353l;
              long fac[]=new long[n+1];
              fac[0]=1l;
              for(int i=1;i<=n;i++)
              {
                  fac[i]=(fac[i-1]*(long)i)%m;
              }
              mergeSort(a,0,n-1);
              int dif=a[n-1]-a[n-2];
              if(dif>1)
              {
                  out.println(0);
              }
              else if(dif==1)
              {
                  int c=0;
                  for(int i=n-2;i>=0;i--)
                  {
                      if(a[n-1]-a[i]==1)
                      {
                          c++;
                      }
                      else
                      {
                          break;
                      }
                  }
                  long sum=0;
                  for(int i=c+1;i<=n;i++)
                  {
                      long p1=fac[c];
                      long p2=comb(fac,i-1,c,m);
                      long p3=fac[n-c-1];
                      long p=(p1*p2)%m;
                      p=(p*p3)%m;
                      sum=(sum+p)%m;
                  }
                  long ans=(fac[n]-sum)%m;
                  if(ans<0) ans+=m;
                  out.println(ans);
              }
              else
              {
                  out.println(fac[n]);
              }
          }
          out.close();
      }
      public static long comb(long fac[],int nu,int b,long m)
      {
          long x=fac[nu];
          long y=(fac[b]%m*fac[nu-b]%m)%m;
          long v=power(y,m-2,m);
          long z=(x%m*v%m)%m;
          return z;
      }
      public static long power(long a,long b,long m)
      {
          long res=1;
          while(b>0)
          {
              if(b%2!=0)
              {
                  res=(res%m*a%m)%m;
              }
              b=b/2;
              a=(a%m*a%m)%m;
          }
          return res;
      }
      public static void mergeSort(int a[],int p,int r)
      {
          if(p<r)
          {
              int q=(p+r)/2;
              mergeSort(a,p,q);
              mergeSort(a,q+1,r);
              merge(a,p,q,r);
          }
      }
      public static void merge(int a[],int p,int q,int r)
      {
          int n1=q-p+2;
          int L[]=new int[n1];
          int n2=r-q+1;
          int R[]=new int[n2];
          for(int i=p;i<=q;i++)
          {
              L[i-p]=a[i];
          }
          L[n1-1]=Integer.MAX_VALUE;
          for(int i=q+1;i<=r;i++)
          {
              R[i-q-1]=a[i];
          }
          R[n2-1]=Integer.MAX_VALUE;
          int x=0,y=0;
          for(int i=p;i<=r;i++)
          {
              if(L[x]<=R[y])
              {
                  a[i]=L[x];
                  x++;
              }
              else
              {
                  a[i]=R[y];
                  y++;
              }
          }
      }
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
          public FastReader()
          {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          int nextInt()
          {
              return Integer.parseInt(next());
          }
          long nextLong()
          {
              return Long.parseLong(next());
          }
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
          String nextLine()
          {
              String str="";
              try
              {
                  str=br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }