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.util.*;
import java.io.*;

public class Jury_Meeting {

	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;
		}

	}

	public static void shuffle(long[] a) {
		Random r = new Random();

		for (int i = 0; i <= a.length - 2; i++) {
			int j = i + r.nextInt(a.length - i);

			swap(a, i, j);
		}

		Arrays.sort(a);
	}

	public static void swap(long[] a, int i, int j) {
		long temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader t = new FastReader();
		PrintWriter o = new PrintWriter(System.out);
		int test = t.nextInt();
		long mod = 998244353;
		long[] fac = new long[200001];
		fac[0] = 1;

		for (int i = 1; i < fac.length; ++i) {
			fac[i] = (fac[i - 1] * i) % mod;
		}

		while (test-- > 0) {
			int n = t.nextInt();
			long[] a = new long[n];

			for (int i = 0; i < n; ++i)
				a[i] = t.nextLong();

			shuffle(a);

			if (a[n - 1] == a[n - 2])
				o.println(fac[n]);
			else if (a[n - 1] >= a[n - 2] + 2)
				o.println("0");
			else {
				long k = 0;

				for (long v : a)
					if (a[n - 2] == v)
						++k;

				long nCr = fun(n, k + 1, mod, fac);
				long invalid = ((nCr * fac[(int) k]) % mod) * fac[(int) (n - k - 1)] % mod;
				long ans = (fac[n] - invalid + mod) % mod;

				o.println(ans);
			}
		}

		o.flush();
		o.close();
	}

	private static long power(long x, long y, long p) {
		long res = 1;
		x = x % p;

		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x) % p;

			y = y >> 1;
			x = (x * x) % p;
		}

		return res;
	}

	private static long modInverse(long n, long p) {
		return power(n, p - 2, p);
	}

	private static long fun(long n, long r, long mod, long[] fac) {
		if (n < r)
			return 0;

		if (r == 0)
			return 1;

		return (fac[(int) n] * modInverse(fac[(int) r], mod) % mod * modInverse(fac[(int) (n - r)], mod) % mod) % mod;
	}
}