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.*;

public class Main {

    static long mod = 998244353;

    static long binp(long a, long p) {
        long ans = 1;
        while (p > 0) {
            if (p % 2 == 1)
                ans = ans * a % mod;
            a = a * a % mod;
            p >>= 1;
        }
        return ans;
    }

    static long divide_mod(long a, long b) {
        return a * binp(b, mod - 2) % mod;
    }

    static long fact(int a) {
        long x = 1;
        for (int i = 2; i <= a; i++) {
            x = (x * i) % mod;
        }
        return x;
    }

    public static void main(String[] args) throws IOException {
        //br = new BufferedReader(new FileReader("input.txt"));
        //out = new PrintWriter("output.txt");
        int t = nextInt();
        while (t-- != 0) {
            int n = nextInt();
            Long[] a = new Long[n];
            long max1 = -1;
            long max2 = -1;
            TreeSet<Long> s = new TreeSet<>();
            TreeMap<Long, Integer> tm = new TreeMap<>();
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
                s.add(a[i]);
                tm.put(a[i], tm.getOrDefault(a[i], 0) + 1);
                if (max1 < a[i]) {
                    max2 = max1;
                    max1 = a[i];
                } else if (max2 < a[i] && a[i] != max1) max2 = a[i];
            }
            if (max2 == -1) out.println(fact(n));
            else {
                if (tm.get(max1) == 1 && max1 - max2 > 1) out.println(0);
                else {
                    long ans = 0;
                    if (tm.get(max1) > 1) {
                        ans=fact(n);
                    } else {
                        long z = 1;
                        for (int i = tm.get(max2) + 2; i < n + 1;++i)
                            z = z * i % mod;
                        ans = fact(tm.get(max2)) * tm.get(max2) % mod *z % mod;
                    }
                    out.println(ans);
                }
            }
        }
        ;
        out.close();
    }

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer in = new StringTokenizer("");

    public static boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    public static String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}

class Ayo implements Comparable<Ayo> {
    long a;
    long b;

    Ayo(long a, long b) {
        this.a = a;
        this.b = b;
    }

    public int compareTo(Ayo o) {
        if (o.a == a) return -Long.compare(b, o.b);
        else return Long.compare(a, o.a);
    }
}