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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.*;
 
public class JuryMeeting {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       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 long[] fact;
    public static long m = 998244353;
    
    public static void find()
    {
    	fact[0] = 1;
    	for(int i = 1; i<=200000; i++)
    	   fact[i] = (fact[i-1] * i)%m;
    	   
    }
	
	public static long solution(int[] arr, int n)
			
	{
		int max1 = Integer.MIN_VALUE;
		
		for(int i = 0; i<n; i++)
		{
			if(arr[i]>max1)
				max1 = arr[i];
		}
		
        int max2 = Integer.MIN_VALUE;
		
		for(int i = 0; i<n; i++)
		{
			if(arr[i]>max2 && arr[i]<max1)
				max2 = arr[i];
		}
		
		if(max2 == Integer.MIN_VALUE)
			return fact[n];
		
		int num  = 0;
		int c = 0;
		
		for(int i = 0; i<n; i++)
		{
			if(arr[i]==max2)
				num++;
			if(arr[i]==max1)
				c++;
		}
		
		long ans = 0;
		
		if(c>1)
			return fact[n];
		
		if(max1-max2>1)
			return 0;
		
		for(int i = num; i<n; i++)
			ans = (ans + ((fact[i]/fact[i-num])*(fact[n-num-1]))%m)%m;

		return (fact[n] - ans + m)%m;
			
	}
        
private static PrintWriter out = new PrintWriter(System.out);

public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
    fact = new long[200001];
    find();
    
    int t = s.nextInt();
    
    for(int j = 0; j<t ; j++)
    {
    	int n = s.nextInt();
    	int[] arr = new int[n];
    	for(int i =0; i<n; i++)
    		arr[i] = s.nextInt();
    	
    	out.println(solution(arr,n));
    }
    
    out.flush();
    out.close();
    
}
}
