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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class TaskC {
    final static long mod = 998244353L;

    public static void main(String[] arg) {
        final FastScanner in = new FastScanner(System.in);
        final PrintWriter out = new PrintWriter(System.out);
        final int T = in.nextInt();
        for (int t = 0; t < T; t++) {
            final int n = in.nextInt();
            final int a[] = in.readIntArr(n);
            final long sol = solution(n, a);
            out.println(sol);
        }
        out.flush();
        out.close();
        in.close();
    }

    private static long solution(int n, int[] a) {

        final int max = max(a);
        int[] cntMap = new int[3];
        for (int i = 0; i < n; i++) {
            if (a[i] == max) {
                cntMap[2]++;
                continue;
            }
            if (a[i] + 1 == max) {
                cntMap[1]++;
                continue;
            }
            cntMap[0]++;
        }
        if (cntMap[2] >= 2) {
            return fact(n);
        }
        if (cntMap[1] == 0) {
            return 0;
        }
        long res = fact(n);
        for (int maxPos = n; maxPos-1 >= cntMap[1]; maxPos--) {
            long minus = fact(maxPos - 1) * combinations(cntMap[0], n - maxPos);
            res -= minus;
        }
        return res;
    }

    final static long combinations(final long n, final long k) {
        return fact(n) / fact(n - k);
    }

    final static int max(final int[] a) {
        int max = a[0];
        for (int i = 1; i < a.length; i++) {
            if (max < a[i]) {
                max = a[i];
            }
        }
        return max;
    }

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


    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

        int[] readIntArr(int n) {
            int[] result = new int[n];
            for (int i = 0; i < n; i++) {
                result[i] = Integer.parseInt(next());
            }
            return result;
        }

        long[] readLongArr(int n) {
            long[] result = new long[n];
            for (int i = 0; i < n; i++) {
                result[i] = Long.parseLong(next());
            }
            return result;
        }


        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

        long nextLong() {
            return Long.parseLong(next());
        }

    }


}