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 C_Jury_Meeting{
    public static void main(String[] args) {
        FastScanner s= new FastScanner();
        StringBuilder res = new StringBuilder();
        long mod=998244353;
        int size=(int)Math.pow(10,5);
        size=2*size;
        size+=5;
        long factorial[]= new long[size];
        factorial[1]=1;
        factorial[0]=1;
        for(int i=2;i<size;i++){
            long f=i*factorial[i-1];
            f=f%mod;
            factorial[i]=f;
        }
        int t=s.nextInt();
        int p=0;
        while(p<t){
        int n=s.nextInt();
        ArrayList<Long> list = new  ArrayList<Long>();
        for(int i=0;i<n;i++){
            long a=s.nextLong();
            list.add(a);
        }
        
        Collections.sort(list);
        long a=list.get(n-1);
        long b=list.get(n-2);
        if(a==b){
            long ans=factorial[n];
           
            res.append(ans+"\n");
        }
        else if(b+1==a){
            int count=0;
            long num=b;
            for(int i=0;i<n;i++){
                if(list.get(i)==num){
                    count++;
                }
            }
            int remain=n-count-1;
            long ans=0;
            for(int i=n-1;i>=count;i--){
                int here=i;
                int there=n-i-1;
                int choose=here-count;
                long yo=(factorial[here]*factorial[there])%mod;
                long a1=factorial[remain];
                long a2=factorial[remain-choose];
                long a3=factorial[choose];
                long a22=modInverse(a2,mod);
                long a33=modInverse(a3,mod);
                long yo1=(a1*a22)%mod;
                yo1=(yo1*a33)%mod;
                yo=(yo*yo1)%mod;
                ans+=yo;
                ans=ans%mod;





            }
            long comp=factorial[n];
            comp=comp-ans+mod;
            comp=comp%mod;
            res.append(comp+"\n");



        }
        else{
            res.append("0\n");
        }
        p++;
    }
    System.out.println(res);
    }
    static long power(long x, long y, long m)
    {
        if (y == 0)
            return 1;
        long p = power(x, y / 2, m) % m;
        p = (long)((p * (long)p) % m);
        if (y % 2 == 0)
            return p;
        else
            return (long)((x * (long)p) % m);
    }
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    static long modInverse(long a, long m)
    {
        long g = gcd(a, m);
       return power(a, m - 2, m);
        
        
            // If a and m are relatively prime, then modulo
            // inverse is a^(m-2) mode m
           
        
    }
    static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
 
		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
 
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(nextToken());
		}
 
		long nextLong() {
			return Long.parseLong(nextToken());
		}
 
		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}




}