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 c {
    
    public static long MODVAL = 998244353;
    
    public static long dp[] = new long[100001];
    
    public static long perm(int a, int b){
        
        return (fact(a)/fact(a-b))%MODVAL;
        
    }
    
    public static long fact(int x){
        
        if(dp[x] != 0){
            return dp[x];
        }
        
        return dp[x] = (x*(fact(x-1))%MODVAL)%MODVAL;
        
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        dp[0] = 1;
        dp[1] = 1;
        while(t-- > 0){

            int n = sc.nextInt();
            long arr[] = new long[n];
            long maxVal = -1;
            long secondMax = -2;
            for(int i=0; i<n; i++){
                arr[i] = sc.nextInt();
                if(arr[i] > maxVal){
                    secondMax = maxVal;
                    maxVal = arr[i];
                }
                else if(arr[i] > secondMax && arr[i] < maxVal){
                    secondMax = arr[i];
                }
            }

            int countMax = 0;
            int countSecondMax = 0;
            for(int i=0; i<n; i++){
                if(arr[i] == secondMax){
                    countSecondMax++;
                }
                if(arr[i] == maxVal){
                    countMax++;
                }
            }
            
            if(secondMax < 0) {
                System.out.println(fact(n));
            }
            else if(maxVal - secondMax > 1 && countMax != n){
                System.out.println(0);
            }
            else if(countMax > 1){
                System.out.println(fact(n));
            }
            else{
                
                int c = n-countMax-countSecondMax;
                
                long ans = fact(n);
                
                ans = (ans - (countMax*fact(n-1))%MODVAL)%MODVAL;
                
                for(int i=1; i<=c; i++){
                    
                    ans = (ans - (countMax * (fact(n-i-1) * perm(c, i))%MODVAL)%MODVAL)%MODVAL;
                    
                }
                
                System.out.println(ans);
                
            }

        }

        sc.close();
    }

}
