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 C {
    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        for(int a=0;a<t;a++) {
            int m = sc.nextInt();
//            sc.nextLine();
            int mod = 998244353;
//            String[] inp = sc.nextLine().split(" ");
            Integer[] tasks = new Integer[m];
            for(int i=0;i<m;i++){
                tasks[i] = sc.nextInt();
            }
            Collections.shuffle(Arrays.asList(tasks));
 
            Arrays.sort(tasks);
            if(tasks[m-1]-tasks[m-2]>1){
                System.out.println(0);
            } else if (tasks[m-1]-tasks[m-2]==1) {
                int idx = m-2;
                int cnt=0;
                while( idx>=0 &&tasks[idx].equals(tasks[m-2])) {
                    cnt++; idx--;
                }
                long res = 1;
                for (int i = 1; i <= m; i++) {
                    if(i==(cnt+1)) continue;
                    res = (res * i) % mod;
                }
//                int res = modFact((m),mod);
//                double last = ((cnt/(cnt+1)));
//                System.out.println(last);
                res = (res * cnt) % mod;
 
                System.out.println(res);
            } else {
                System.out.println(modFact(m,mod));
            }
        }
    }
 
    static int modFact(int n, int p)
    {
        long result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;
 
        return (int) result;
    }
 
    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;
        }
    }
}
 