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 {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
  	static String readLine() throws IOException {
  		return br.readLine();
  	}
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(readLine());
  		return st.nextToken();
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readChar() throws IOException {
  		return next().charAt(0);
  	}
  	static class Pair implements Comparable<Pair> {
  		int f, s;
  		Pair(int f, int s) {
  			this.f = f; this.s = s;
  		}
  		public int compareTo(Pair other) {
  			if (this.f != other.f) return this.f - other.f;
  			return this.s - other.s;
  		}
  	}
  	final static int N = 200000, mod = 998244353;
  	static long fac[] = new long[N + 1], inv[] = new long[N + 1];
  	static long pow(long b, long p) {
  		long ret = 1;
  		while (p > 0) {
  			if ((p&1) == 1) ret = b*ret%mod;
  			b = b*b%mod;
  			p >>= 1;
  		}
  		return ret;
  	}
  	static long C(int n, int k) {
  		return fac[n]*(inv[k]*inv[n - k]%mod)%mod;
  	}
  	static void solve() throws IOException {
  		int n = readInt(), max1 = 0, max2 = 0, a[] = new int[n + 1];
  		for (int i = 1; i <= n; ++i) {
  			a[i] = readInt();
  			fac[i] = i*fac[i - 1]%mod;
  			if (a[i] > max1) {
  				max2 = max1;
  				max1 = a[i];
  			} else if (a[i] > max2) max2 = a[i];
  		}
  		if (max1 - max2 >= 2) {
  			pr.println(0);
  			return;
  		}
  		if (max1 - max2 == 0) {
  			pr.println(fac[n]);
  			return;
  		}
  		int cnt = 0;
  		for (int i = 1; i <= n; ++i)
  			if (a[i] >= max2) ++cnt;
  		long bad = C(n, cnt);
  		bad = bad*fac[cnt - 1]%mod;
  		bad = bad*fac[n - cnt]%mod;
  		pr.println((fac[n] - bad + mod)%mod);
  	}
  	public static void main(String[] args) throws IOException {
  		//solve();
  		fac[0] = inv[0] = 1;
  		for (int i = 1; i <= N; ++i) {
  			fac[i] = i*fac[i - 1]%mod;
  			inv[i] = pow(fac[i], mod - 2);
  		}
  		for (int t = readInt(); t > 0; --t) solve();
  		pr.close();
  	}
  }
  