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 {

    static long fact[] = new long[200005];
    static long M = (long)998244353;

    public static long bin_exp(long a, long b){

        if(b == 0){
            return 1;
        }

        long x = bin_exp(a, b/2);

        if(b%2 == 0){
            long ans = (x*x)%M;
            return ans;
        }

        else{
            long ans = a*(x*x%M)%M;
            return ans;
        }

    }

    public static long inv(long a){

        return bin_exp(a, M-2);

    }

    public static long perm(int a, int b){

        return (fact[a] * inv(fact[a-b]))%M;

    }

    public static long binom(int a, int b){

        if (b < 0 || a < 0)return 0;
        return (fact[a] * ((inv(fact[b]) * inv(fact[a-b]))%M)%M)%M;

    }

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        fact[0] = 1L;
        for(int i=1; i<fact.length; i++){
            fact[i] = (fact[i-1]*i)%M;
        }


        while(t-- > 0){

            int n = sc.nextInt();
            int arr[] = new int[n];

            for(int i=0; i<n; i++){
                arr[i]  =sc.nextInt();
            }

            int maxVal = arr[0];
            int secondMaxVal = -1;
            for(int i=1; i<n; i++){
                if(arr[i] > maxVal){
                    secondMaxVal = maxVal;
                    maxVal = arr[i];
                }
                else if(arr[i] > secondMaxVal){
                    secondMaxVal = arr[i];
                }
            }

            if(maxVal == secondMaxVal){
                System.out.println(fact[n]);
                continue;
            }

            if(maxVal - secondMaxVal > 1){
                System.out.println(0);
                continue;
            }

            int count = 0;
            for(int i=0; i<n; i++){

                if(arr[i] == secondMaxVal){
                    count++;
                }

            }


            long ans = fact[n];
            ans = (ans - fact[n-1]);
            if(ans < 0){
                ans += M;
            }
            ans = ans%M;
            int iter = n-count-1;

            long sum = 0L;
            for(int i=1; i<=iter; i++){
                sum = (sum + (perm(iter, i) * fact[n-1-i])%M)%M;
            }

            ans  = ans - sum;
            if(ans < 0){
                ans += M;
            }
            ans = ans%M;

            System.out.println(ans);

        }

    }

}