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.*;
public class new1{
	
	
	public static long fact(int n, int m, int mod) {
		long fact = 1;
		for(int i = 2; i <= n; i++) {
			if(i != m) {
				fact = (fact * i) % mod;
			}
		}
		return fact;
	}

    
    public static void main(String[] args)  throws IOException{
    	FastReader s = new FastReader();
    	BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
    	int t = s.nextInt();
    	for(int i = 0; i < t; i++) {
    		int n = s.nextInt();
    		long[] arr = new long[n];
    		ArrayList<Integer> aList = new ArrayList<Integer>();
    		for(int j = 0; j < n; j++) {
    			aList.add(s.nextInt());
    		}
    		Collections.sort(aList, null);
    		for(int j = 0; j < n; j++) {
    			arr[j] = aList.get(j);
    		}
    		if(arr[n - 1] - arr[n - 2] >= 2) {
    			System.out.println(0);
    			continue;
    		}
    		long ans = -1;
    		int mod = 998244353;
    		if(arr[n - 1] == arr[n - 2]) ans = fact(n, 1, mod);
    		else {
    			int x = 1;
    			for(int j = n - 3; j >= 0; j--) {
    				if(arr[j] == arr[j + 1]) x++;
    				else break;
    			}
    			//System.out.println(fact(n, 1, mod) + " " + fact(n, x, mod));
    			ans = (fact(n, x + 1, mod) * x) % mod;
    		}
    		System.out.println(ans);
    		
    		
    		
    		output.flush();
    		
    	}
            
        
        
    }
}
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();
    }
 
    public 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;
}}