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.io.*;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Random;
import java.util.StringTokenizer;

public class Main {
    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);


    static long[] factorial = new long[200001];
    static int mod =  998244353;

    static void precomputeFactorial(){
        factorial[0] = 1;
        for(int i = 1; i < factorial.length; i++){
            factorial[i] = (factorial[i-1] * i) % mod;
        }
    }

    public static void main(String[] args) throws IOException {

        precomputeFactorial();
        int t = sc.nextInt();

        while(t-- > 0){
            int n = sc.nextInt();
            long[] A = new long[n];

            HashMap<Long, Integer> map = new HashMap<>();
            for(int i = 0 ; i < n ; i++){
                A[i] = sc.nextLong();
                if(!map.containsKey(A[i])){
                    map.put(A[i], 0);
                }
                map.put(A[i], map.get(A[i]) + 1);
            }

            Arrays.sort(A);

            long max = A[n - 1];
            long secondMax = A[n - 2];

            if(max - secondMax > 1){
                out.println(0);
            }else if(max - secondMax == 0){
                out.println(factorial[n]);
            }else{
                int k = map.get(secondMax);
                long ans = (factorial[n] + mod - (factorial[n] / (k + 1)) % mod) % mod;
                out.println((int)ans);
            }


        }




        out.flush();
        out.close();
    }



    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }


        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.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 = reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return str;
        }
    }

    public static void shuffle(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }

    public static void shuffle(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }


}
