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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int i,N=200000;
        long f[]=new long[N+1],in[]=new long[N+1],M=998244353;
        f[0]=1;
        for(i=1;i<=N;i++) f[i]=f[i-1]*i%M;
        in[N]=power(f[N],M-2,M);
        for(i=N-1;i>=0;i--) in[i]=in[i+1]*(i+1)%M;

        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine());
            int a[]=new int[n];
            String s[]=bu.readLine().split(" ");

            PriorityQueue<Integer> pq=new PriorityQueue<>();
            for(i=0;i<n;i++)
            {
                a[i]=Integer.parseInt(s[i]);
                pq.add(a[i]);
            }
            for(i=0;i<n;i++) a[i]=pq.poll();

            long ans=0;
            if(a[n-1]-a[n-2]>1) ans=0;
            else if(a[n-1]==a[n-2]) ans=f[n];
            else
            {
                int secl=0;
                for(i=0;i<n;i++)
                if(a[i]==a[n-2]) secl++;

                secl=n-(secl+1);
                for(i=0;i<n;i++)
                if(secl>=n-1-i)
                ans=(ans+ncr(f,in,secl,n-1-i)*f[n-1-i]%M*f[i]%M)%M;
                ans=(f[n]-ans+M)%M;
            }
            sb.append(ans+"\n");
        }
        System.out.print(sb);
    }

    static long ncr(long f[],long in[],int n,int r)
    {
        if(n<r) return 0;
        long M=998244353;
        return f[n]*in[r]%M*in[n-r]%M;
    }

    static long power(long a,long b,long M)
    {
        long res=1;
        while(b!=0)
        {
            if(b%2==1) res=res*a%M;
            b>>=1;
            a=a*a%M;
        }
        return res;
    }
}