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.
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 int MOD = 998244353;
	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();
		int fact[] = new int[2 * 100000 + 3];
		fact[0] = 1;
		for (int i = 1; i <= 2 * 100000 + 2; i++) {
			fact[i] = fact[i - 1] * i;
			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)
						right = fact[n - i - 1] % MOD * ncr(n - countSecond - 1, n - i - 1, fact) % MOD;

					if (i == n - 1)
						right = 1;

					long left = fact[i];
					long sub = right % MOD * left % MOD;
					ans -= sub;

				}
				pw.println(ans);

			} 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, int[] 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());
	}
}