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.ArrayList;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class Solve {
  
  	static long MOD = 998244353L;
  	static long INF = 1000_000_000_000_000L + 100;
  	static boolean isPrime[];
  
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner();
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		long fact[] = new long[3 * 100000 + 3];
  		fact[0] = 1L;
  		for (int i = 1; i <= 3 * 100000 + 2; i++) {
  			fact[i] = (fact[i - 1] % MOD * i * 1L) % MOD;
  			fact[i] %= MOD;
  
  		}
  
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int a[] = sc.nextIntArray(n);
  			sort(a);
  			int max = a[n - 1];
  			if (a[n - 1] - 1 > (a[n - 2])) {
  				pw.println(0);
  				continue;
  			}
  
  			int countMax = 0;
  			for (int i : a) {
  				if (i == max)
  					countMax++;
  			}
  
  			if (countMax == 1) {
  				long ans = fact[n];
  				int countSecond = 0;
  				for (int i : a)
  					if (i == max - 1)
  						countSecond++;
  
  				for (int i = n - 1; i >= 0; i--) {
  					long right = 0L;
  					if (n - countSecond - 1 >= n - i - 1) {
  						long x = (fact[n - i - 1] % MOD);
  						long y = ncr(n - countSecond - 1, n - i - 1, fact) % MOD;
  						right = (x * 1L * y) % MOD;
  					}
  					if (i == n - 1)
  						right = 1L;
  
  					long left = fact[i] * 1L;
  
  					long sub = (right % MOD * (left % MOD)) % MOD;
  
  					ans -= sub;
  					ans %= MOD;
  
  				}
  				pw.println((ans + MOD) % MOD);
  
  			} else {
  				pw.println(fact[n]);
  			}
  
  		}
  		pw.flush();
  
  	}
  
  	static long LowerBound(int a[], int x, int start) { // x is the target value or key
  		int l = start, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] >= x)
  				r = m;
  			else
  				l = m;
  		}
  		return r;
  	}
  
  	static long UpperBound(int a[], int x, int start) {// x is the key or target value
  		int l = start, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] <= x)
  				l = m;
  			else
  				r = m;
  		}
  		return l + 1;
  	}
  
  	static public boolean[] sieve(int n) {
  		boolean prime[] = new boolean[n + 1];
  		for (int i = 0; i <= n; i++)
  			prime[i] = true;
  
  		for (int p = 2; p * p <= n; p++) {
  			// If prime[p] is not changed, then it is a
  			// prime
  			if (prime[p] == true) {
  				// Update all multiples of p
  				for (int i = p * p; i <= n; i += p)
  					prime[i] = false;
  			}
  		}
  		return prime;
  
  	}
  
  	public static ArrayList<Integer> primeFactors(int n) {
  		ArrayList<Integer> set = new ArrayList<Integer>();
  
  		// Print the number of 2s that divide n
  		while (n % 2 == 0) {
  			n /= 2;
  			set.add(2);
  		}
  
  		// n must be odd at this point. So we can
  		// skip one element (Note i = i +2)
  		for (int i = 3; i * i <= n; i += 2) {
  			// While i divides n, print i and divide n
  			while (n % i == 0) {
  				n /= i;
  				set.add(i);
  			}
  		}
  
  		if (n > 1)
  			set.add(n);
  
  		return set;
  
  	}
  
  	static int gcd(int a, int b) {
  		if (a == 0)
  			return b;
  		if (b == 0)
  			return a;
  
  		return gcd(b, a % b);
  
  	}
  
  	public static int[] swap(int data[], int left, int right) {
  
  		// Swap the data
  		int temp = data[left];
  		data[left] = data[right];
  		data[right] = temp;
  
  		// Return the updated array
  		return data;
  	}
  
  	public static long[] euclidExtended(long a, long b) {
  		if (b == 0) {
  			long ar[] = new long[] { 1, 0 };
  			return ar;
  		}
  		long ar[] = euclidExtended(b, a % b);
  		long temp = ar[0];
  		ar[0] = ar[1];
  		ar[1] = temp - (a / b) * ar[1];
  		return ar;
  	}
  
  	public static long modInverse(long a, long m) {
  		long ar[] = euclidExtended(a, m);
  		return ((ar[0] % m) + m) % m;
  	}
  
  	// Fermat Theorem
  	public static long ncr(int n, int r, long[] fact) {
  		if (n == r)
  			return 1;
  		if (r == 1)
  			return n;
  		if (n == 0 || r == 0)
  			return 1;
  		return (fact[n] * modInverse(((long) fact[n - r]) * fact[r], MOD)) % MOD;
  	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l = new ArrayList<>();
  		for (int i : a)
  			l.add(i);
  		Collections.sort(l);
  		for (int i = 0; i < a.length; i++)
  			a[i] = l.get(i);
  	}
  
  }
  
  class Pair {
  	int x, y;
  
  	Pair(int x, int y) {
  		this.x = x;
  		this.y = y;
  	}
  
  }
  
  class SegTree {
  	int leftmost, rightmost, sum;
  	SegTree lChild, rChild;
  
  	public SegTree(int leftmost, int rightmost) {
  		this.leftmost = leftmost;
  		this.rightmost = rightmost;
  	}
  
  	public void update(int index, int val) {
  		sum++;
  		if (leftmost == rightmost) {
  			return;
  		}
  		if (lChild == null)
  			makeKids();
  		if (index <= lChild.rightmost)
  			lChild.update(index, val);
  		else
  			rChild.update(index, val);
  	}
  
  	public void makeKids() {
  		int mid = (leftmost + rightmost) / 2;
  		lChild = new SegTree(leftmost, mid);
  		rChild = new SegTree(mid + 1, rightmost);
  	}
  
  	public int query(int l, int r) {
  		if (l <= leftmost && r >= rightmost)
  			return sum;
  		if (l > rightmost || r < leftmost)
  			return 0;
  		if (lChild == null)
  			return 0;
  		return lChild.query(l, r) + rChild.query(l, r);
  	}
  }
  
  class UnionFind {
  	int[] p, rank, setSize;
  	int numSets;
  
  	public UnionFind(int N) {
  		p = new int[numSets = N];
  		rank = new int[N];
  		setSize = new int[N];
  		for (int i = 0; i < N; i++) {
  			p[i] = i;
  			setSize[i] = 1;
  		}
  	}
  
  	public int findSet(int i) {
  		return p[i] == i ? i : (p[i] = findSet(p[i]));
  	}
  
  	public boolean isSameSet(int i, int j) {
  		return findSet(i) == findSet(j);
  	}
  
  	public void unionSet(int i, int j) {
  		if (isSameSet(i, j))
  			return;
  		numSets--;
  		int x = findSet(i), y = findSet(j);
  		if (rank[x] > rank[y]) {
  			p[y] = x;
  			setSize[x] += setSize[y];
  		} else {
  			p[x] = y;
  			setSize[y] += setSize[x];
  			if (rank[x] == rank[y])
  				rank[y]++;
  		}
  	}
  
  	public int numDisjointSets() {
  		return numSets;
  	}
  
  	public int sizeOfSet(int i) {
  		return setSize[findSet(i)];
  	}
  }
  
  class Scanner {
  	BufferedReader br;
  	StringTokenizer st;
  
  	Scanner() {
  		br = new BufferedReader(new InputStreamReader(System.in));
  	}
  
  	String next() throws IOException {
  		while (st == null || !st.hasMoreElements())
  			st = new StringTokenizer(br.readLine());
  		return st.nextToken();
  	}
  
  	int nextInt() throws NumberFormatException, IOException {
  		return Integer.parseInt(next());
  	}
  
  	int[] nextIntArray(int n) throws NumberFormatException, IOException {
  		int a[] = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = Integer.parseInt(next());
  		return a;
  	}
  
  	long nextLong() throws NumberFormatException, IOException {
  		return Long.parseLong(next());
  	}
  
  	long[] nextLongArray(int n) throws NumberFormatException, IOException {
  		long a[] = new long[n];
  		for (int i = 0; i < n; i++)
  			a[i] = Long.parseLong(next());
  		return a;
  	}
  
  	double nextDouble() throws NumberFormatException, IOException {
  		return Double.parseDouble(next());
  	}
  }