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.*;
public class Jury{
	private static Scanner sc=new Scanner(System.in);
	public static long mod=998244353;
    public static long fact(long n){
    	long res=1;
    	for(int i=1;i<=n;i++)
    		res=(res%mod*i%mod)%mod;
    	return res;
    }
	public static  void solve(){
        int n=sc.nextInt();
        long a[]=new long[n];
        for(int i=0;i<n;i++){
        	a[i]=sc.nextInt();
        }
        long ma=0;
       for(int i=0;i<n;i++){
       	ma=Math.max(a[i],ma);
       }
       int count=0,count1=0;
       for(int i=0;i<n;i++)
       	{if(a[i]==ma)count++;
       	 if(a[i]==ma-1)count1++;
       	}
       	if(count>=2){System.out.println(fact(n));}
       	else{
       		long sub=1;
       		for(int i=1;i<=n;i++){
       			if(i!=count1+1)sub=(sub*i)%mod;
       		}
       		System.out.println((fact(n)-sub)%mod);
       	}
	}
	public static void main(String args[]){
		
		int t=sc.nextInt();
		while(t-->0){
			solve();
		}
	}
}