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.io.PrintWriter;
  import java.util.HashMap;
  import java.util.StringTokenizer;
  
  
  
  public class JuryMeeting {
  
      static long mod=998244353;
  
      public static void main(String[] args) {
  
          FastScanner f = new FastScanner();
          PrintWriter p = new PrintWriter(System.out);
          long fac[]=new long[200005];
          fac[0]=fac[1]=1;
          for(int i=2;i<fac.length;i++){
              fac[i]=mul(i,fac[i-1]);
          }
          int t = f.nextInt();
          while (t-- > 0) {
              int n=f.nextInt();
              long max=Long.MIN_VALUE;
              HashMap<Long,Long> h=new HashMap<>();
              for(int i=0;i<n;i++){
                  long curr=f.nextLong();
                  if(h.containsKey(curr)){
                      h.put(curr,h.get(curr)+1);
                  }else{
                      h.put(curr,(long)1);
                  }
                  max=Math.max(max,curr);
              }
  
              if(h.get(max)>1){
                  p.println(fac[n]);
              }else if(h.get(max)==1){
                  if(h.containsKey(max-1)==false){
                      p.println(0);
                  }else{
                      long k=h.get(max-1);
                      long ans=sub(fac[n],div(fac[n],k+1));
                      p.println(ans);
                  }
              }
  
          }
          p.close();
  
      }
  
  //-----------------------------------------------------------------------------------------------------------------------------------------------------------------
  
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner() {
              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;
          }
      }
  
      static long add(long x,long y){
          x += y;
          if(x >= mod) return (x % mod);
          return x;
      }
      static long sub(long x, long y){
          x -= y;
          if(x < 0)   return (x + mod);
          return x;
      }
      static long mul(long x, long y){
          return (x * y) % mod;
      }
      static long bin_pow(long x,long p){
          if(p == 0)  return 1;
          if((p & 1)!=0)   return mul(x, bin_pow(x, p - 1));
          return bin_pow(mul(x,x), p / 2);
      }
      static long rev(long  x) {
          return bin_pow(x, mod - 2);
      }
      static long div(long x, long y){
          return mul(x, rev(y));
      }
  
  
      // will work till array size of 65537---fastest sorting time
  //	static long[] radixSort(long[] f) {
  //		return radixSort(f, f.length);
  //	}
  //
  //	static long[] radixSort(long[] f, int n) {
  //		long[] to = new long[n];
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 16 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 16 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 32 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 32 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		{
  //			int[] b = new int[65537];
  //			for (int i = 0; i < n; i++)
  //				b[1 + (int) (f[i] >>> 48 & 0xffff)]++;
  //			for (int i = 1; i <= 65536; i++)
  //				b[i] += b[i - 1];
  //			for (int i = 0; i < n; i++)
  //				to[b[(int) (f[i] >>> 48 & 0xffff)]++] = f[i];
  //			long[] d = f;
  //			f = to;
  //			to = d;
  //		}
  //		return f;
  //	}
  
      // Function to sort an array using quick sort algorithm.
      static void quickSort(long arr[], int low, int high) {
          if (low < high) {
              int p = partition(arr, low, high);
              quickSort(arr, low, p - 1);
              quickSort(arr, p + 1, high);
          }
      }
  
      static int partition(long arr[], int low, int high) {
          long pivot = arr[high];
          int i = low - 1;
          for (int j = low; j < high; j++) {
              if (arr[j] < pivot) {
                  i++;
                  long temp = arr[i];
                  arr[i] = arr[j];
                  arr[j] = temp;
              }
          }
          long temp = arr[i + 1];
          arr[i + 1] = arr[high];
          arr[high] = temp;
  
          return i + 1;
      }
  
      /// Ceils
      static long ceil(long x, long m) {
          long res = x / m;
          if (x % m != 0) {
              res++;
          }
          return res;
      }
  
      // ------------------------------------------------------------------------------------------------
      // makes the prefix sum array
      static long[] prefixSum(long arr[], int n) {
          long psum[] = new long[n];
          psum[0] = arr[0];
          for (int i = 1; i < n; i++) {
              psum[i] = psum[i - 1] + arr[i];
          }
          return psum;
      }
      // ------------------------------------------------------------------------------------------------
  
      // makes the suffix sum array
      static long[] suffixSum(long arr[], int n) {
          long ssum[] = new long[n];
          ssum[n - 1] = arr[n - 1];
          for (int i = n - 2; i >= 0; i--) {
              ssum[i] = ssum[i + 1] + arr[i];
          }
          return ssum;
      }
  
  //------------------------------------------------------------------------------------------
      // BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
      // OVERHEADS
  
      static long binPower(long a, long n, long m) {
          if (n == 0)
              return 1;
          long res = 1;
          while (n > 0) {
              if ((n & 1) != 0) {
                  res *= a;
              }
              a *= a;
              n >>= 1;
          }
          return res;
      }
  
  //-------------------------------------------------------------------------------------------
  
      // gcd
      static long gcd(long a, long b) {
          return b == 0 ? a : gcd(b, a % b);
      }
  
  //------------------------------------------------------------------------------------------
  
      // lcm
      static long lcm(long a, long b) {
          return a / gcd(a, b) * b;
      }
  
      //------------------------------------------------------------------------------------------
      // BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
      // O(LOGn)
      static int setBits(int n) {
          int count = 0;
          while (n > 0) {
              n = n & (n - 1);
              count++;
          }
          return count;
      }
  //------------------------------------------------------------------------------------------
  
      //------------------------------------------------------------------------------------------
      // 0 based indexing
      static boolean KthBitSet(int n, int k) {
          int mask = 1;
          mask = mask <<= k;
          if ((mask & n) != 0)
              return true;
          else
              return false;
      }
  
  //------------------------------------------------------------------------------------------
  
      // EXTENDED EUCLIDEAN THEOREM
      // TO REPRESENT GCD IN TERMS OF A AND B
      // gcd(a,b) = a.x + b.y where x and y are integers
  
      static long x = -1;
      static long y = -1;
  
      static long gcdxy(long a, long b) {
          if (b == 0) {
              x = 1;
              y = 0;
              return a;
          } else {
              long d = gcdxy(b, a % b);
              long x1 = y;
              long y1 = x - (a / b) * y;
              x = x1;
              y = y1;
  
              return d;
          }
      }
  
  //-------------------------------------------------------------------------------------------------F
  }
  