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.Scanner;
public class Solution {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
           Scanner sc=new Scanner (System.in);
           int t =sc.nextInt();
           while(t-->0) {
        	int n=sc.nextInt();
        	long max =-1;
        	long []arr =new long[n];
        	for(int i=0;i<n;i++) {
        		arr[i]=sc.nextLong();
        		max =Math.max(max, arr[i]);
        	}
        	 int count1=0;
        	 int count2=0;
        	 
        	 
        	 for(int i=0;i<n;i++) {
        		 if(arr[i]==max)count1++;
        		 
        		 if(arr[i]==max-1)count2++;
        	 }
        	 if(count1>1) {
        		 System.out.println(fact(n,998244353));
        	 }else if(count2>0) {
        		 System.out.println(fact(n,998244353) - sub(n ,998244353,count2));
        	 }else {
        		 System.out.println("0");
        	 }
        	 
           }
           
}
	  static int fact (int n ,long mod ) {
		 long ans=1;
		  for(int i=1;i<=n;i++) {
			  ans=ans*i;
		  }
		  return (int)(ans%mod);
	  }
	  static int sub(int n , long mod,int count2) {
		  long sub=1;
		  for(int i=2;i<=n;i++) {
			  if(i!=count2+1) {
				  sub=sub*i;
			  }
			 
		  }
		  return (int)(sub%mod);
	  }
}