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.Scanner; 
import java.util.StringTokenizer;
import java.util.*;
public class codeforcesA{
   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; 
        } 
   } 
   public static void main(String args[]){
     FastReader sc=new FastReader();
     int t=sc.nextInt();
     StringBuilder sb=new StringBuilder();
     while(t-->0){
       int n=sc.nextInt();
       int ar[]=new int[n];
       long mod=998244353;
       Map<Integer,Integer> map=new HashMap<>();
       long fact_n=1,fact_n_=1;
       int max=0,min=0;
       for(int i=0;i<n;i++){ar[i]=sc.nextInt();map.put(ar[i],map.getOrDefault(ar[i],0)+1);
          if(ar[i]>=max){min=max;max=ar[i];}else{min=Math.max(ar[i],min);}
        }
       ar[n-2]=min;ar[n-1]=max;
       for(int i=1;i<=n;i++){fact_n=(fact_n*(long)i)%mod;
          if(i!=map.get(ar[n-2])+1){fact_n_=(fact_n_*(long)i)%mod;}
        }long ans=0;
       if(ar[n-1]==ar[n-2]){
         ans=fact_n;
        }
        else{
            if(ar[n-1]-ar[n-2]==1){
             ans=(fact_n-fact_n_+mod)%mod;
            }
            else{
                ans=0;
            }
        }
        sb.append(ans+"\n");
     }
     System.out.print(sb.toString());
    }
}