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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;


public class Main {
    static int i, j, k, n, m, t, y, x, sum = 0;
    static long mod =  998244353;
    static FastScanner fs = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
    static String str;
    static long ans;


    public static void main(String[] args) {

        t = fs.nextInt();
        long[] fact = new long[200005];
        fact[0]=1L;

        for(i=1;i<200005;i++){
            fact[i] = fact[i-1]*i;
            fact[i]%=mod;

        }

        while(t-- >0) {

            n = fs.nextInt();

            int[] arr = new int[n];
            int max=0, nextMax=0, maxCount = 0, nextMaxCount=0;

            for(i=0;i<n;i++){
                arr[i]=fs.nextInt();
                max = Math.max(max,arr[i]);
            }

            for(i=0;i<n;i++){
                if(arr[i]!=max)
                    nextMax = Math.max(nextMax,arr[i]);
            }

            for(i=0;i<n;i++){
                if(arr[i] == max)
                    maxCount++;
                else if(arr[i] == nextMax)
                    nextMaxCount++;
            }

            if(maxCount ==1 && (nextMax == 0 || nextMax!=max-1 ))
                out.println(0);

            else{
                if(maxCount>1)
                    out.println(fact[n]);

                else{

                    int x = n-(maxCount+nextMaxCount);
                    ans = fact[n];

                    long diff = 1;
                    int j=n-1;
                    while(x>=0){


                        diff%=mod;

                        long temp = (diff*fact[j--])%mod;

                        ans-=temp;

                        if(ans<0)
                            ans+=mod;

                        ans%=mod;

                        diff*=(x--);

                    }

                    out.println(ans);
                }

            }



        }

        out.close();
    }

    static long computePow(int x, int y){
        long ans = 1;
        for(i=0;i<y;i++){
            ans*=x;
            ans%=mod;
        }

        return ans;
    }


    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static void ruffleSort(int[] a) {
        //ruffle
        int n = a.length;
        Random r = new Random();
        for (int i = 0; i < a.length; i++) {
            int oi = r.nextInt(n), temp = a[i];
            a[i] = a[oi];
            a[oi] = temp;
        }

        //then sort
        Arrays.sort(a);
    }

    static void ruffleSort(long[] a) {
        //ruffle
        int n = a.length;
        Random r = new Random();
        for (int i = 0; i < a.length; i++) {
            int oi = r.nextInt(n);
            long temp = a[i];
            a[i] = a[oi];
            a[oi] = temp;
        }

        Arrays.sort(a);
    }

    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());
        }
    }

    static class Pair {
        public int x, y;

        Pair(int x, int y){
            this.x = x;
            this.y = y;
        }
    }


}