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: //This code is written by प्रविण शंखपाळ 
  
  //package wizard;
  
  import java.io.*;
  import java.math.*;
  import java.util.*;
  
  import java.text.*;
  
  public class Ginny_Weasley {
  
  	static long mod = (long) 998244353;
  
  	public static void main(String[] args) {
  
  //		try {
  
  		FastReader fr = new FastReader();
  		PrintWriter pt = new PrintWriter(System.out);
  
  		int t = fr.nextInt();
  
  		while (t > 0) {
  
  			int n = fr.nextInt();
  
  			long A[] = new long[n];
  			for (int i = 0; i < n; i++) {
  				A[i] = fr.nextLong();
  			}
  
  			sort(A, 0, n - 1);
  
  			long fac = 1, ans = 0, k = 0, fac1 = 1;
  
  			for (long i = 1; i <= (long) n; i++) {
  				fac *= i;
  				fac %= mod;
  			}
  
  			for (int i = 0; i < n - 1; i++) {
  				if (A[i] == A[n - 1] - 1) {
  					k++;
  				}
  			}
  
  			for (long i = 1; i <= (long) n; i++) {
  				if (i != k + 1) {
  					fac1 *= i;
  					fac1 %= mod;
  				}
  			}
  
  			if (A[n - 1] == A[n - 2]) {
  				ans = fac;
  			} else if (A[n - 1] - A[n - 2] >= 2) {
  				ans = 0;
  			} else {
  
  				ans = (((fac1 * k) % mod));
  				ans %= mod;
  
  			}
  
  			pt.println(ans);
  
  			t--;
  		}
  		pt.close();
  
  //		} catch (
  //
  //		Exception e) {
  //			return;
  //		}
  	}
  
  	static int lower_bound(long[] arr, int begin, int end, long tar) {
  		while (begin < end) {
  			int mid = begin + (end - begin) / 2;
  			if (arr[mid] < tar)
  				begin = mid + 1;
  			else if (arr[mid] >= tar)
  				end = mid;
  		}
  		return begin;
  //		lower bound gives largest index of a number smaller than (if target is not present) or equal to target(if target is present) 
  //      for complete array use begin = 0 and end = n
  	}
  
  	static int upper_bound(long[] arr, int begin, int end, long tar) {
  		while (begin < end) {
  			int mid = begin + (end - begin) / 2;
  			if (arr[mid] <= tar)
  				begin = mid + 1;
  			else if (arr[mid] > tar)
  				end = mid;
  		}
  
  		return begin;
  //		upper bound gives smallest index of a number strictly greater than target
  //	    for complete array use begin = 0 and end = n
  	}
  
  	public static void mergeSort(String[] nameGo) {
  		if (nameGo.length > 1) {
  			String[] leftGo = new String[nameGo.length / 2];
  			String[] rightGo = new String[nameGo.length - nameGo.length / 2];
  			for (int so = 0; so < leftGo.length; so++) {
  				leftGo[so] = nameGo[so];
  			}
  			for (int ki = 0; ki < rightGo.length; ki++) {
  				rightGo[ki] = nameGo[ki + nameGo.length / 2];
  			}
  			mergeSort(leftGo);
  			mergeSort(rightGo);
  			merges(nameGo, leftGo, rightGo);
  		}
  	}
  
  	public static void merges(String[] nameH, String[] leftH, String[] rightH) {
  		int as = 0;
  		int bs = 0;
  		for (int i = 0; i < nameH.length; i++) {
  			if (bs >= rightH.length || (as < leftH.length && leftH[as].compareToIgnoreCase(rightH[bs]) < 0)) {
  				nameH[i] = leftH[as];
  				as++;
  			} else {
  				nameH[i] = rightH[bs];
  				bs++;
  			}
  		}
  	}
  
  	static void merge(long arr[], int l, int m, int r) {
  
  		int n1 = m - l + 1;
  		int n2 = r - m;
  
  		long L[] = new long[n1];
  		long R[] = new long[n2];
  
  		for (int i = 0; i < n1; ++i)
  			L[i] = arr[l + i];
  		for (int j = 0; j < n2; ++j)
  			R[j] = arr[m + 1 + j];
  
  		int i = 0, j = 0;
  
  		int k = l;
  		while (i < n1 && j < n2) {
  			if (L[i] <= R[j]) {
  				arr[k] = L[i];
  				i++;
  			} else {
  				arr[k] = R[j];
  				j++;
  			}
  			k++;
  		}
  
  		while (i < n1) {
  			arr[k] = L[i];
  			i++;
  			k++;
  		}
  
  		while (j < n2) {
  			arr[k] = R[j];
  			j++;
  			k++;
  		}
  	}
  
  	static void sort(long arr[], int l, int r) {
  		if (l < r) {
  
  			int m = l + (r - l) / 2;
  
  			sort(arr, l, m);
  			sort(arr, m + 1, r);
  
  			merge(arr, l, m, r);
  		}
  	}
  
  	static class Pair implements Comparable<Pair> {
  		int a, b;
  
  		Pair(int a, int b) {
  			this.a = a;
  			this.b = b;
  		}
  
  		public int compareTo(Pair o) {
  			// TODO Auto-generated method stub
  			if (this.a != o.a)
  				return Integer.compare(this.a, o.a);
  			else
  				return Integer.compare(this.b, o.b);
  			// return 0;
  		}
  
  		public boolean equals(Object o) {
  			if (o instanceof Pair) {
  				Pair p = (Pair) o;
  				return p.a == a && p.b == b;
  			}
  			return false;
  		}
  
  	}
  
  	static int binarySearch(int arr[], int first, int last, int key) {
  		int mid = (first + last) / 2;
  		while (first <= last) {
  			if (arr[mid] < key) {
  				first = mid + 1;
  			} else if (arr[mid] == key) {
  				return mid;
  			} else {
  				last = mid - 1;
  			}
  			mid = (first + last) / 2;
  		}
  		return -1;
  	}
  
  	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;
  		}
  	}
  }
  