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.*;
import java.io.*;

public class Main {
    // For fast input output
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            try {
                br = new BufferedReader(new FileReader("input.txt"));
                PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                System.setOut(out);
            } catch (Exception e) {
                br = new BufferedReader(new InputStreamReader(System.in));
            }
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.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 = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    // end of fast i/o code
    static int MOD = 998244353;
    public static void main(String[] args) {
        FastReader reader = new FastReader();

        int t = reader.nextInt();
        int n;
        StringBuilder sb = new StringBuilder("");
        long ans = 0;
        int i;
        int max_limit = 2*100000+1;
        long factorial[] = new long[max_limit];
        long inverse[] = new long[max_limit];
        long factorial_inv[] = new long[max_limit];
        fact(factorial, inverse, factorial_inv, max_limit);

        while (t-- > 0) {
            n = reader.nextInt();
            int arr[] = new int[n];
            for(i=0; i<n; i++){
                arr[i] = reader.nextInt();
            }

            // System.out.println(n+"<--> "+factorial[n]);
            ans = factorial[n];

            
            int max = Integer.MIN_VALUE;
            for(i=0; i<n; i++){
                max = Math.max(max, arr[i]);
            }
            
            int max_count = 0;
            for(i=0; i<n; i++){
                if(arr[i]==max){
                    max_count++;
                }
            }

            
            if(max_count>1){
                sb.append(ans+"\n");
                continue;
            }


            int considering = 0;
            for(i=0; i<n; i++){
                if(arr[i]<max-1){
                    considering++;
                }
            }

            if(considering==n-max_count){
                sb.append("0\n");
                continue;
            }

            long ans_temp = 0;
            for(i=0; i<=considering; i++){

                ans_temp = (factorial[considering] * factorial_inv[considering-i])%MOD;
                ans_temp = (ans_temp * factorial_inv[i])%MOD;
                ans_temp = (ans_temp * factorial[n-i-1])%MOD;
                ans_temp = (ans_temp * factorial[i])%MOD;

                // System.out.println((n-i)+"<-,,->"+factorial_inv[n-i]+" "+ans_temp);
                // System.out.println((((n-i)*inverse[n-i])%MOD));
                
                ans = (ans - ans_temp + MOD)%MOD;
                // System.out.println(ans+"--->"+ans_temp+"\n");
            }

            sb.append(ans+"\n");
        }
        System.out.println(sb);
    }

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

        if(b==0){
            return 1;
        }
        if(b==1){
            return a;
        }
        if((b&1)==1){
            return (a*fast_pow(a, b-1))%MOD;
        }
        return fast_pow((a*a)%MOD, b/2);

    }

    public static void fact(long factorial[], long inverse[], long factorial_inv[], int max){

        long current = 1;
        factorial[0] = 1;
        for(int i=1; i<max; i++){
            current = (current * i)%MOD;
            factorial[i] = current;
        }


        inverse[0] = 1;
        for(int i=1; i<max; i++){
            inverse[i] = fast_pow(i, MOD-2);
        }
        
        factorial_inv[0] = 1;
        current = 1;
        for(int i=1; i<max; i++){
            current = (current * inverse[i])%MOD;
            factorial_inv[i] = current;
        }
    }
}