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.lang.*;
import java.io.*;

public class B{
    static long MOD = 998244353;
    static long[] fact;
    static void getfact(){
        fact[0] = 1;
        for(int i=1;i<=200000;i++){
            fact[i] = (fact[i-1]*i)%MOD;
        }
    }
    static long power(long x, long y){
        long res = 1;
        x = x%MOD;
        while(y>0){
            if( y%2 == 1){
                res = (res*x)%MOD;
            }
            y = y/2;
            x = (x*x)%MOD;
        }
        return res;
    }
    static long modInverse(long a){
        return power(a,MOD-2);
    }
    static long nCr(int n,int r){
        long sum = (long)(fact[n]%MOD);
        sum = (sum*modInverse(fact[r]))%MOD;
        sum = (sum*modInverse(fact[n-r]))%MOD;
        return sum;
    }
    public static void main(String[] args){
        FastScanner fs = new FastScanner();
        int t = fs.nextInt();
        fact = new long[200001];
        getfact();
        while(t>0){
            int n = fs.nextInt();
            long [] arr = fs.readArray(n);
            int largest = 0, second_largest = 0, freq_largest = 0, freq_second_largest = 0;
            for(int i=0;i<n;i++){
                if(arr[i]>largest) {
                    second_largest = largest;
                    freq_second_largest = freq_largest;
                    largest = (int)arr[i];
                    freq_largest = 1;
                    continue;
                }
                else if(arr[i] == largest){
                    freq_largest++;
                }else if(arr[i]> second_largest){
                    second_largest = (int)arr[i];
                    freq_second_largest = 1;
                }else if(arr[i] == second_largest)
                    freq_second_largest++;
            }
//            System.out.println("largest: "+largest+" second_largest: "+second_largest);
            if(freq_largest > 1)
                System.out.println(fact[n]);
            else if(largest - second_largest > 1)
                System.out.println(0);
            else{
                long  curr = fact[n];
                int  cnt = n - 1 - freq_second_largest;
                for(int i=0;i<=cnt;i++){
                    long  sum = (nCr(cnt,i)*(fact[i]*fact[n-i-1])%MOD)%MOD;
                    curr = (curr - sum+MOD)%MOD;
                }
                System.out.println(curr);
            }
            t--;
        }
        return;
    }
    static class FastScanner{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next(){
            while(!st.hasMoreTokens()){
                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());
        }
        long[] readArray(int n){
            long[] a = new long [n];
            for(int i=0;i<n;i++)a[i] = nextLong();
            return a;
        }
    }
}