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.StringTokenizer;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Collections;
import java.util.ArrayList;

public class p3
{
	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()); }

		byte nextByte() { return Byte.parseByte(next()); }

		short nextShort() { return Short.parseShort(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 main(String[] args)throws Exception
	{
		FastReader fr=new FastReader();
		BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));

		short t=fr.nextShort();
		while(t-->0)
		{
			int n=fr.nextInt();
			int a[]=new int[n];
			for(int i=-1;++i<n;)
				a[i]=fr.nextInt();

			//3 4 2 1 3 3
			//2 3 1 - 2 2
			//1 2 - - 1 1
			//

			int max=0;
			int maxc=0;
			int max2=0;
			int max2c=0;
			int mod=998244353;
			for(int i=-1;++i<n;)
			{
				if(a[i]>max)
				{
					max2=max;
					max2c=maxc;
					max=a[i];
					maxc=0;
				}
				if(a[i]==max)
					maxc++;

				if(a[i]==max2)
					max2c++;				
			}

			if(maxc>1)
			{
				long ans=1;
				for(int i=0;++i<=n;)
				{
					ans*=i;
					ans%=mod;
				}
				System.out.println(ans);
			}
			else if(max-max2>1)
			{
				System.out.println(0);
			}
			else
			{
				long ans=1;
				for(int i=0;++i<=n;)
				{
					if(i==max2c+1)
					{
						ans*=max2c;
						ans%=mod;
						continue;
					}
					ans*=i;
					ans%=mod;
				}
				System.out.println(ans);
			}
		}
	}
}