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.math.*;
import java.util.*;


// @author : Dinosparton 

public class test {
	 
	   static class Pair{ 
		   long x;
		   long y;
		   
		   Pair(long x,long y){ 
			   this.x = x;
			   this.y = y;
			   
		   }
	   }
	  
	   static class Sort implements Comparator<Pair>
	   {

	       @Override
	       public int compare(Pair a, Pair b)
	       {
	           if(a.x!=b.x)
	           {
	               return (int)(a.x - b.x);
	           }
	           else
	           {
	               return (int)(a.y-b.y);
	           }
	       }
	   }
	   
	   static class Compare { 
		   
		     void compare(Pair arr[], int n) 
		    { 
		        // Comparator to sort the pair according to second element 
		        Arrays.sort(arr, new Comparator<Pair>() { 
		            @Override public int compare(Pair p1, Pair p2) 
		            { 
		            	if(p1.x!=p2.x) {
		                return (int)(p1.x - p2.x); 
		            	}
		            	else { 
		            		return (int)(p1.y - p2.y);
		            	}
		            } 
		        }); 
		  
//		        for (int i = 0; i < n; i++) { 
//		            System.out.print(arr[i].x + " " + arr[i].y + " "); 
//		        } 
//		        System.out.println(); 
		    } 
		} 
	 
	   static class Scanner {
	        BufferedReader br;
	        StringTokenizer st;
	 
	        public Scanner()
	        {
	            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;
	        }
	    }
	 
	   static long mod = 998244353;
	   static long mul(long x, long y){
	        return (x * y) % mod;
	    }
	   static long bin_pow(long x,long p){
	        if(p == 0)  return 1;
	        if((p & 1)!=0)   return mul(x, bin_pow(x, p - 1));
	        return bin_pow(mul(x,x), p / 2);
	    }
	    static long rev(long  x) {
	        return bin_pow(x, mod - 2);
	    }
	    static long div(long x, long y){
	        return mul(x, rev(y));
	    }
	    static long add(long x,long y){
	        x += y;
	        if(x >= mod) return (x % mod);
	        return x;
	    }
	    static long sub(long x, long y){
	        x -= y;
	        if(x < 0)   return (x + mod);
	        return x;
	    }
	    
	   public static void main(String args[]) throws Exception { 
		
		   
		   Scanner sc = new Scanner();
		   StringBuilder res = new StringBuilder();
		
		   long fac[]=new long[200005];
	        fac[0]=fac[1]=1;
	        for(int i=2;i<fac.length;i++){
	            fac[i]=mul(i,fac[i-1]);
	        }
	        
		   int tc = sc.nextInt();
		   
		   while(tc-->0) { 
			
			   int n = sc.nextInt();
	            long max=Long.MIN_VALUE;
	            
	            HashMap<Long,Long> map=new HashMap<>();
	            
	            for(int i=0;i<n;i++){
	            	
	                long curr=sc.nextLong();
	                
	                if(map.containsKey(curr)){
	                    map.put(curr,map.get(curr)+1);
	                }else{
	                    map.put(curr,(long)1);
	                }
	                
	                max=Math.max(max,curr);
	            }
	 
	            if(map.get(max)>1){
	                res.append(fac[n]+"\n");
	            }else if(map.get(max)==1){
	                if(map.containsKey(max-1)==false){
	                    res.append(0+"\n");
	                }else{
	                    long k=map.get(max-1);
	                    long ans=sub(fac[n],div(fac[n],k+1));
	                    res.append(ans+"\n");
	                }
	            }
	 
		   }
			System.out.println(res);
		   
	   }
}  




