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.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public class A {

    public static void main(String[] args) {
        FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        int t = scan.nextInt();
        for(int tt = 1; tt <= t; tt++) solver.solve(tt, scan, out);
        out.close();
    }

    static class Task {
        static long[] fact, invFact, naturalInverse;
        static final long MOD = 998244353;
        static final int MAX = (int) (2e5 + 100);

        public void solve(int testNumber, FastReader scan, PrintWriter out) {
            if(testNumber == 1) precomp();

            int n = scan.nextInt();
            int[] a = new int[n];

            for(int i = 0; i < n; i++) a[i] = scan.nextInt();
            ruffleSort(a);
            if(a[n - 1] == a[n - 2]) out.println(fact[n]);
            else if(a[n - 1] > a[n - 2] + 1) out.println(0);
            else {
                long ans = 0;
                int count = 0;

                for(int i = 0; i < n; i++) {
                    if(a[i] == a[n - 2]) count++;
                }
                for(int i = 0; i < n; i++) {
                    long total = fact[n - 1];
                    long bad = perm(i, count) * fact[n - 1 - count] % MOD;
                    total = ((total - bad) % MOD + MOD) % MOD;
                    ans = (ans + total) % MOD;
                }
                out.println(ans);
            }
        }

        void precomp() {
            fact = new long[MAX + 1];
            invFact = new long[MAX + 1];
            naturalInverse = new long[MAX + 1];

            fact[0] = invFact[0] = invFact[1] = naturalInverse[0] = naturalInverse[1] = 1;
            for(int i = 1; i <= MAX; i++) {
                fact[i] = (fact[i-1]*i)%MOD;
                if(i == 1) continue;
                naturalInverse[i] = naturalInverse[(int) (MOD % i)] * (MOD - MOD/i) % MOD;
                invFact[i] = (invFact[i-1]*naturalInverse[i])%MOD;
            }
        }

        static long perm(int a, int b) {
            if(a < b) return 0;
            return(fact[a]%MOD*invFact[a-b])%MOD;
        }

    }

    static void ruffleSort(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
        Arrays.sort(a);
    }

    static void ruffleSort(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
        Arrays.sort(a);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                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());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}