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.*;
  import java.io.*;
  public class Sol{
   
  /*   
  ->check n=1, int overflow , array bounds , all possibilites(dont stuck on 1 approach)
  ->Problem = Observation(constraints(m<=n/3 or k<=min(100,n)) 
             + Thinking + Technique (seg_tree,binary lift,rmq,bipart,dp,connected comp etc)
  ->solve or leave it (- tutorial improves you in minimal way -)
  */
   
  public static void main (String []args) {
  precomp();
  
  int times=ni();while(times-->0){solve();}out.close();}
   
  static void solve(){
    int n=ni();
    HashMap<Integer,Integer> hs=new HashMap<>();int max=0;
    for(int i=1;i<=n;i++){int x=ni();
      if(hs.containsKey(x)){hs.put(x,hs.get(x)+1);}
      else{hs.put(x,1);}
      max=Math.max(max,x);}
     
    if(hs.get(max)>1){out.println(factorial[n]);}
    else if(hs.containsKey(max-1)){
       int cnt=hs.get(max-1)+1;
       long ans=factorial[n];
       long delta=(C(n,cnt)*factorial[cnt-1])%mod;
       delta=(delta*factorial[n-cnt])%mod;
       ans=(ans+mod-delta)%mod;
       out.println(ans);
    }
    else{out.println(0);}
    return;
  }
   
  
  //-----------------Utility--------------------------------------------
   
  static int N=2000005;
  static long inv[];
  static long factorial[];
   
  static void precomp(){
   inv=new long[N+1];factorial=new long[N+1];
   inv[0]=1;factorial[0]=1l;
   long j=1;
   for(int i=1;i<=N;i++){factorial[i]=(j*factorial[i-1])%mod;j++;}
   j=N-1;
   inv[N]=power(factorial[N],mod-2);
   for(int i=N-1;i>=0;i--){ inv[i]=((j+1)*inv[i+1])%mod;j--;}
   
  }
   
  static long C(int n,int r){
     if(r<0 || r>n)return 0l;
     long a=( ( ( factorial[n]*inv[r] )%mod )*inv[n-r])%mod;
     return a;
   }
  
  
  static long gcd(long a,long b){if(b==0)return a; return gcd(b,a%b);}
   
  static int Max=Integer.MAX_VALUE; static long mod=998244353;
   
  //static int v(char c){return (int)(c-'a');}
   
  public static long power(long x, long y )
      {
          //0^0 = 1
          long res = 1L;
          x = x%mod;
          while(y > 0)
          {
              if((y&1)==1)
                  res = (res*x)%mod;
              y >>= 1;
              x = (x*x)%mod;
          }
          return res;
      }
   
  /*static class Pair implements Comparable<Pair>{
          int id;int value;Pair next;
          public Pair(int id,int value) {
           
            this.id=id;this.value=value;next=null;
          }
          @Override
          public int compareTo(Pair p){return Long.compare(value,p.value);}
      }
  */ 
   
  //----------------------I/O---------------------------------------------
   
  static InputStream inputStream = System.in;
  static OutputStream outputStream = System.out;
  static FastReader in=new FastReader(inputStream);
  static PrintWriter out=new PrintWriter(outputStream);
   
  static class FastReader 
  { 
    BufferedReader br; 
    StringTokenizer st; 
   
          FastReader(InputStream is) {
              br = new BufferedReader(new InputStreamReader(is));
          }
   
    public String next() 
    { 
        while (st == null || !st.hasMoreElements()) 
        { 
            try
            { 
                st = new StringTokenizer(br.readLine()); 
            } 
            catch (IOException  e) 
            { 
                e.printStackTrace(); 
            } 
        } 
        return st.nextToken(); 
    } 
   
    public int nextInt() 
    { 
        return Integer.parseInt(next()); 
    } 
   
    public long nextLong() 
    { 
        return Long.parseLong(next()); 
    } 
   
   public  double nextDouble() 
    { 
        return Double.parseDouble(next()); 
    } 
   
   
    String nextLine() 
    { 
        String str = ""; 
        try
        { 
            str = br.readLine(); 
        } 
        catch (IOException e) 
        { 
            e.printStackTrace(); 
        } 
        return str; 
    } 
  }
  //Only for Integer Input
  /*static int ni() {
          try {
              boolean in = false;
              int res = 0;
              for (;;) {
                  int b = System.in.read() - '0';
                  if (b >= 0) {
                      in = true;
                      res = 10 * res + b;
                  }
                  else if (in)
                      return res;
              }
          } catch (IOException e) {
              throw new Error(e);
          }
      }*/
  static int ni(){return in.nextInt();}
  static long nl(){return in.nextLong();}
  static double nd(){return in.nextDouble();}
  static String ns(){return in.nextLine();}
  }