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.*;
import static java.lang.System.out;
import java.util.Stack;
import java.util.Queue;

public class C1569
{

    static int mod=998244353;
    static long MOD=998244353L;
    static FastReader in=new FastReader();
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static int fact[] = new int[(int)1e5+5]; 
    public static void main(String args[])
    {  

        int tc=1;
        factorial();
        tc=in.nextInt();
        tcloop: while(tc-->0)
        {
            
            int n=in.nextInt();
            int arr[] = in.readIntArray(n);

            sort(arr);
            HashMap<Integer,Integer> hm = new HashMap<>();
        
            int max = 0;
            for(int i : arr){

                max = Math.max(i, max);
                hm.put(i,hm.getOrDefault(i, 0)+1);

            }
            if(hm.get(max)>=2){
                pr.println(fact[max]);
            }
            else{
                if(!hm.containsKey(max-1)){
                    pr.println(0);
                }
                else{
                    
                    int val = hm.get(max-1);

                    int ans = fact[n] - fact[n]/(val+1);
                    pr.println(ans);

                }
            }
          
            
            
            


        }
        pr.flush();
        
    }

    static void factorial(){
        fact[0]=1;
        for(int i = 1;i<(int)1e5+5;i++){
            fact[i] = (i*fact[i-1])%mod;
        }
    }

	static long gcd(long a,long b)
	{
		if(a==0)return b;
		return gcd(b%a,a);
	}

    static class Pair implements Comparable<Pair>
    {
        int a,b;
        Pair(int a,int b)
        {
            this.a=a;
            this.b=b;

        }

        @Override
        public int compareTo(Pair o)
        {
            return Integer.compare(a,o.a);
        }
    }

    
    static void dsort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
        Collections.reverse(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
    

    

    static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
    
    
	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
    }

    
    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()); 
        } 
        
        int[] readIntArray(int n)
        {
		int a[]=new int[n];
		for(int i=0;i<n;i++)a[i]=nextInt();
		return a;
	}
		
	long[] readLongArray(int n)
	{
		long a[]=new long[n];
		for(int i=0;i<n;i++)a[i]=nextLong();
		return a;
	}
	
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
}



