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.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;


public class JuryMeeting {

    private static final long MOD=998244353;
    private static InputStreamReader ina = new InputStreamReader(System.in);
    private static BufferedReader in = new BufferedReader(ina);

    public static void main(String[] args )throws Exception {

        int teste = Integer.parseInt(in.readLine());
        while (teste-- > 0) {
            solve();
        }
    }

    private static void solve() throws Exception {
        int inimesi = Integer.parseInt(in.readLine());
        String[] tasks_str = in.readLine().split(" ");
        ArrayList<Long> tasks = new ArrayList<>();
        long max_tasks = 0;
        for (int i = 0; i < inimesi; i++) {
            long task = Integer.parseInt(tasks_str[i]);
            if(task > max_tasks){
                max_tasks = task;
            }
            tasks.add(task);
        }
        int max_taskide_arv = Collections.frequency(tasks, max_tasks);
        int max_miinus_1_taskide_arv = Collections.frequency(tasks, max_tasks-1);
        int ylejaanud = tasks.size() - max_taskide_arv - max_miinus_1_taskide_arv;
        // Kõigil sama arv taske st kõik on max_tasks
        if(max_miinus_1_taskide_arv == 0 && ylejaanud == 0){
            System.out.println(permutatsioonid(max_taskide_arv));
        } else if (max_miinus_1_taskide_arv == 0){ // Kui on muid ka siis peab leiduma vähemnalt 1 inimene kellel on taske max_taskide_arv - 1
            System.out.println("0");
        } else {
            // Kõik võimalikud
            long permutatsioonid = permutatsioonid(tasks.size());

            // Nüüd hakkame maha lahutama neid mis ei sobi e kus kõik max_miinus_1_taskide_arv on eespool
            // Niipalju kui on ylejäänuuid tuleb kombineerida
            // 3 3 3 4 1 2 kõik max miinus-1 ees
            // 3 3 3 1 4 2 kõik max miinus-1 1 ees ja 1 ylejäänud ka
            // 3 3 3 1 2 4
            for (int i = 0; i <= ylejaanud; i++) {

                // Kõik miinus-1 on eesotsas ja lisaks ylejäänutest i
                long eesmise_otsa_perm = permutatsioonid(max_miinus_1_taskide_arv + i);

                // Tuleb arvestada, et maksimume võib mitu olla, siis tekib tagumises otsa
                // permutatsioone ylejaanud -i  korda max_taskide_arv
                long tagumise_otsa_perm = (permutatsioonid(ylejaanud - i) * max_taskide_arv)%MOD;

                // Lisaks kombinatsioonid mida saab teha ylejäänutes koguarvust niipalju kaupa kui on ees otas
                // eesotsas on i tükki, järelikult i kaupa ylejäänust
                long kombinatsioonid = kombinatsioonid(ylejaanud, i);

                // Pane tähele, et vahepeal tuleb %MOD teha, muidu "jookseb üle" KONTROLLI !
                permutatsioonid = (permutatsioonid - ((eesmise_otsa_perm * tagumise_otsa_perm)%MOD * kombinatsioonid)%MOD + MOD) % MOD;
            }
            System.out.println(permutatsioonid);
        }
    }

    private static long kombinatsioonid(long millest, long kaupa) {
        long retVal = 1;
        for (long d = 1; d <= kaupa; ++d) {
            retVal *= millest--;  // Hakkame tagantpoolt, ehk kui 10, siis retval = 1*10 esimene kord, järgmine 10 * 9, kuni kaupa jätkub
            retVal /= d;  // Kuna muurujoone all oli k!(n-k)! siis siin jagame seda k!-d 1, 2, 3  (n-k)! pole vaja sest eelmine samm väldib seda
            retVal %= MOD; // Siin peaks ka tohtima MOD teha
        }
        return retVal;
    }

    private static long permutatsioonid(long max_taskide_arv) {
        long retVal = 1;
        for (int i = 1; i <= max_taskide_arv ; i++) {
         retVal = (retVal*i)%MOD;
        }
        return retVal;
    }
}
