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.*;
import java.util.*;
//solve harder problems
//upsolve
//use gitgud
//learn from everything
//USACO
public class Contest1569C
{
	static class InputReader {
		BufferedReader reader;
		StringTokenizer tokenizer;
		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		String next() { // reads in the next string
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() { // reads in the next int
			return Integer.parseInt(next());
		}
		public long nextLong() { // reads in the next long
			return Long.parseLong(next());
		}
		public double nextDouble() { // reads in the next double
			return Double.parseDouble(next());
			}
		}
	static InputReader r = new InputReader(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
	static long m = 998244353;
	public static void main(String[] args)
	{
		//LOOK FOR INT OVERFLOW
		//LOOK FOR SCOPE ERROR
		//Add negatives
		//SORT BOTH ARRAYS
		//CHECK THE LAST ONE THAT IS OUT LOOP
		//SORT BY BOTH VALUES
		//DON'T ADD STRINGS,USE STRING BUILDER
		int t = r.nextInt();
		while (t > 0)
		{
			t--;
			int n = r.nextInt();
			int[] a = new int[n];
			int max = 0;
			for (int i = 0; i < n; i ++)
			{
				a[i] = r.nextInt();
				max = Math.max(a[i], max);
			}
			int countM = 0;
			int countm = 0;
			for (int i = 0; i < n; i ++)
			{
				if (a[i] == max)
				{
					countM ++;
				}
				else if (a[i] == max - 1)
				{
					countm ++;
				}
			}
			if (countM > 1)
			{
				pw.println(factorial(n));
			}
			else if (countm == 0)
			{
				pw.println(0);
			}
			else
			{
				pw.println((factorial(n)-((factorial(n)*(binpow(countm+1,m-2)%m))%m)+m)%m);
			}
		}
		
		pw.close();
	}
	public static long factorial(long x)
	{
		if (x == 1 || x == 0)
		{
			return 1;
		}
		return ((x%m) * (factorial(x-1)%m))%m;
	}
	public static long binpow(long x, long n) {
		assert(n >= 0);
		x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow
		long res = 1;
		while (n > 0) {
			if (n % 2 == 1)  // if n is odd
				res = res * x % m;
			x = x * x % m;
			n /= 2;  // divide by two
		}
		return res;
	}
}