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 Main {
    public static void main(String args[])
    {
        FastReader input=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int T=input.nextInt();
        while(T-->0)
        {
            int n=input.nextInt();
            int a[]=new int[n];
            for(int i=0;i<n;i++)
            {
                a[i]=input.nextInt();
            }
            long m=998244353;
            long fac[]=new long[n+1];
            fac[0]=1;
            for(int i=1;i<=n;i++)
            {
                fac[i]=(fac[i-1]*(long)i)%m;
            }
            mergeSort(a,0,n-1);
            int dif=a[n-1]-a[n-2];
            if(dif>1)
            {
                out.println(0);
            }
            else if(dif==1)
            {
                int c=0;
                for(int i=n-2;i>=0;i--)
                {
                    if(a[n-1]-a[i]==1)
                    {
                        c++;
                    }
                    else
                    {
                        break;
                    }
                }
                long sum=0;
                for(int i=c+1;i<=n;i++)
                {
                    long p1=fac[c];
                    long p2=comb(fac,i-1,c,m);
                    long p3=fac[n-c-1];
                    long p=(p1*p2)%m;
                    p=(p*p3)%m;
                    sum=(sum+p)%m;
                }
                long ans=(fac[n]-sum)%m;
                out.println(ans);
            }
            else
            {
                out.println(fac[n]);
            }
        }
        out.close();
    }
    public static long comb(long fac[],int nu,int b,long m)
    {
        long x=fac[nu];
        long y=(fac[b]%m*fac[nu-b]%m)%m;
        long v=power(y,m-2,m);
        long z=(x%m*v%m)%m;
        return z;
    }
    public static long power(long a,long b,long m)
    {
        long res=1;
        while(b>0)
        {
            if(b%2!=0)
            {
                res=(res%m*a%m)%m;
            }
            b=b/2;
            a=(a%m*a%m)%m;
        }
        return res;
    }
    public static void mergeSort(int a[],int p,int r)
    {
        if(p<r)
        {
            int q=(p+r)/2;
            mergeSort(a,p,q);
            mergeSort(a,q+1,r);
            merge(a,p,q,r);
        }
    }
    public static void merge(int a[],int p,int q,int r)
    {
        int n1=q-p+2;
        int L[]=new int[n1];
        int n2=r-q+1;
        int R[]=new int[n2];
        for(int i=p;i<=q;i++)
        {
            L[i-p]=a[i];
        }
        L[n1-1]=Integer.MAX_VALUE;
        for(int i=q+1;i<=r;i++)
        {
            R[i-q-1]=a[i];
        }
        R[n2-1]=Integer.MAX_VALUE;
        int x=0,y=0;
        for(int i=p;i<=r;i++)
        {
            if(L[x]<=R[y])
            {
                a[i]=L[x];
                x++;
            }
            else
            {
                a[i]=R[y];
                y++;
            }
        }
    }
    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;
        }
    }
}