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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class CF1 {
    public static void main(String[] args) {
        FastScanner sc=new FastScanner();
        int T=sc.nextInt();

        for (int tt=0; tt<T; tt++) {
            long n = sc.nextInt();
            long arr[]= new long [(int) n];
            long max=Integer.MIN_VALUE;
            long countmax=0;
            for (int i =0; i<n ; i++){
                arr[i]=sc.nextLong();
                if (arr[i]>max) {
                    max=arr[i];
                    countmax=1;
                }
                else if (arr[i]==max) countmax++;
            }
            if (countmax>1) {
                System.out.println(factorial(n)%mod);
            }
            else {
                long max2=Integer.MIN_VALUE;
                long countmax2=0;
                for (int i =0; i<n ; i++){
                    if (arr[i]==max) continue;
                    else if (arr[i]>max2) {
                        max2=arr[i];
                        countmax2=1;
                    }
                    else if (arr[i]==max2) countmax2++;
                }
                if (max2==max-1) {
                    long wrong=0;
                    for (int i=(int)countmax2; i<=n-1; i++){
                        wrong+= choose(i, countmax2)* factorial(countmax2)* factorial(n-countmax2-1);
                    }
                    long ans = factorial(n)- wrong;
                    System.out.println(ans%mod);
                }
                else System.out.println(0);
            }
        }
    }
    static long mod = 998244353;
    static long factorial(long a){
        if (a==0|| a==1) return 1;
        else return (a*factorial(a-1))%mod;
    }
    static long choose (long a , long b){
        return factorial(a)/(factorial(b)*factorial(a-b));
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}

