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.
// package codeforce.Training1900;

import java.io.PrintWriter;
import java.util.*;

public class JuryMeeting {
    //    MUST SEE BEFORE SUBMISSION
//    check whether int part would overflow or not, especially when it is a * b!!!!
//    check if top down dp would cause overflow or not !!!!!!!!!!!!!!!!!!!!!!
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = 1;
        t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }
    static int mod = 998244353;
    static Combinatorial cb = new Combinatorial(mod, (int) 2e5 + 5);
    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        Map<Integer, Integer> mp = new HashMap<>();
        int max = 0;
        for(int x : a) {
            mp.put(x, mp.getOrDefault(x, 0) + 1);
            max = Math.max(max, x);
        }
        if (mp.get(max) == 1 && mp.getOrDefault(max - 1, 0) == 0){
            out.println(0);
        }else{
            long tot = 1;
            for(int i = n; i >= 1; i--){
                tot = (tot * i) % mod;
            }
            long[] pre = new long[n + 1];
            pre[1] = 1;
            for(int i = 2; i <= n; i++){
                pre[i] = (pre[i - 1] * i) % mod;
            }
            if (mp.get(max) >= 2){
                out.println(tot);
            }else{
                tot = 1;
                for(int i = n - 1; i >= 1; i--){
                    tot = (tot * i) % mod;
                }
                tot = (tot * (n - 1)) % mod;
                int rest = n - mp.get(max) - mp.get(max - 1);
                long ad = 0;
                for(int i = 1; i <= rest; i++){
                    long right = mut(cb.ncr(rest, i), pre[i]);
                    long left = pre[n - 1 - i];
                    ad = add(ad, mut(left, right));
                }
                tot -= ad;
                if (tot < 0) tot += mod;
                out.println(tot);
            }
        }
    }
    static long add(long a, long b){
        a += b;
        if (a >= mod) a-= mod;
        return a;
    }
    static long mut(long a, long b){
        return (a * b) % mod;
    }
    //      credits to SecondThread
    // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
    // worst case since it uses a version of quicksort. Although this would never
    // actually show up in the real world, in codeforces, people can hack, so
    // this is needed.
    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);
    }
    // credits to cuiaoxiang
    static class Combinatorial {

        int MOD;
        int n;
        long[] fact;
        long[] inv;
        long[] ifact;
        public Combinatorial(int mod, int n) {
            this.MOD = mod;
            this.n = n;
            fact = new long[n + 1];
            inv = new long[n + 1];
            ifact = new long[n + 1];
            inv[1] = 1;
            for (int i = 2; i < n; ++i) {
                inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
            }
            fact[0] = ifact[0] = 1;
            for (int i = 1; i < n; ++i) {
                fact[i] = fact[i - 1] * i % MOD;
                ifact[i] = ifact[i - 1] * inv[i] % MOD;
            }

        }

        long ncr(int n, int m){
            if (n < m || m < 0) return 0;
            return fact[n] * ifact[m] % MOD * ifact[n - m] % MOD;
        }
    }

}
