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 Main {
     
     static long invfact[]=new long[200005];
     static long fact[]=new long[200005];
     static long mod= 998244353;
   public static void precompute(){
    fact[0]=fact[1]=1;
    for(int i=2;i<200005;i++)
    {
        fact[i]=fact[i-1]*(long)i;
        fact[i]%=mod;
    }
   }
  public static long modinv(long k)
{
    return powm(k, mod-2);
}
public static long powm(long x, long y)
    {
        long res = 1;
        x = x % mod;
        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % mod;
             y = y >> 1;
            x = (x * x) % mod;
        }
        return res;
    }
  public static long nCr(int n,int r)
    {
          if (n<r)
              return 0;
        if (r == 0)
            return 1;
        return (fact[n]*modinv(fact[r])%mod*modinv(fact[n - r])%mod)%mod;
    }
     
    public static void main(String[] args) {
        
        PrintWriter out=new PrintWriter(System.out);
        Scanner sc=new Scanner(System.in);
        precompute();
        int t=sc.nextInt();
        while(t-->0) {
            int n=sc.nextInt();
            long a[]=new long[n];
            
            for(int i=0;i<n;i++) a[i]=sc.nextLong();
            sort(a);
            long total=fact[n];
            
          //  out.println(total);
            
            if(a[n-1]==a[n-2]) {
                out.println(total);
               // continue;
            }
            else if((a[n-2]+1)!=a[n-1]) {
                 out.println(0);
               //continue;
            }
            else {
                int countSecMax=0;
                for(int i=0;i<n;i++) {
                    if(a[i]==a[n-2]) countSecMax++;
                }
                for(int i=0;i<n;i++) {
                    if(i<countSecMax) continue;
                    total=(mod+total-nCr(i,countSecMax)%mod*fact[countSecMax]%mod*fact[n-countSecMax-1]%mod)%mod;
                 
                }
                out.println(total);
                
           }
        }
        out.flush();
        out.close();
        
        
    }
    static void sort(long a[]){
        ArrayList<Long> arr=new ArrayList<>();
        for(int i=0;i<a.length;i++)arr.add(a[i]);
        Collections.sort(arr);
        for(int i=0;i<a.length;i++)a[i]=arr.get(i);
    }
}