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.lang.*;
import java.io.*;

//           THIS TEMPLATE MADE BY AKSH BANSAL.

public class Solution {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            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;
        }
    }
    private static boolean[] isPrime;
    private static void primes(){
        int num = (int)1e6; // PRIMES FROM 1 TO NUM
        isPrime = new boolean[num];
     
        for (int i = 2; i< isPrime.length; i++) {
           isPrime[i] = true;
        }
        for (int i = 2; i< Math.sqrt(num); i++) {
           if(isPrime[i] == true) {
              for(int j = (i*i); j<num; j = j+i) {
                 isPrime[j] = false;
              }
           }
        }
    }
    static void sort(int a[]){ // int -> long
        ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
        for(int i=0;i<a.length;i++)
        arr.add(a[i]);
        Collections.sort(arr);
        for(int i=0;i<a.length;i++)
        a[i]=arr.get(i);
        
    }
    private static long gcd(long a, long b){
        if(b==0)return a;
        return gcd(b,a%b);
    }
    private static long pow(long x,long y){
        if(y==0)return 1;
        long temp = pow(x, y/2);
        if(y%2==1){
            return x*temp*temp;
        }
        else{
            return temp*temp;
        }
    }

    static PrintWriter out;
    static FastReader sc ;
    public static void main(String[] args) throws IOException {
        sc = new FastReader();
        out = new PrintWriter(System.out);
        // primes();
        // ________________________________

        int test = sc.nextInt();
        StringBuilder output = new StringBuilder();

        while (test-- > 0) {
            int n = sc.nextInt();
            int[] arr =new int[n];
            for(int i=0;i<n;i++){
                arr[i ] =sc.nextInt();
            }
            output.append(solver(arr,n)).append("\n");
        }

        out.println(output);
        // _______________________________

        // int n = sc.nextInt();
        // out.println(solver());
        // ________________________________
        out.flush();
    }
    static int mod =  998244353;
    public static long solver(int[] arr, int n) {
        sort(arr);
        if(arr[n-1]-arr[n-2]>1){
            return 0;
        }
        int count = 1;
        int x = 1;
        for(int i=n-2;i>=0;i--){
            if(arr[n-1]==arr[i]){
                count++;
                x++;
            }
            else if(arr[n-1]==arr[i]+1){
                x++;
            }
        }
        long res =1;
        // System.out.println("__"+ count);
        for(int i=1;i<=n;i++){
            if(count==1 && i==x){
                res = (res*(i-1))%mod;
                continue;
            }
            res = (res*i)%mod;
        }
        return res;
    }
}