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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeMap;
import java.util.StringTokenizer;
import java.util.Map;
import java.util.Map.Entry;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Khater
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CJuryMeeting solver = new CJuryMeeting();
        solver.solve(1, in, out);
        out.close();
    }

    static class CJuryMeeting {
        long[] fac;

        long sub(long a, long b, long mod) {
            return (a - b + mod) % mod;
        }

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {

            fac = new long[(int) (2e5 + 1)];
            long mod = 998244353;
            for (int i = 0; i < fac.length; i++) {
                if (i == 0) fac[i] = 1;
                else fac[i] = (fac[i - 1] * i) % mod;
            }
//		pw.println(Arrays.toString(fac));

            int t = 1;
            t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                int[] arr = sc.nextIntArr(n);
                TreeMap<Integer, Integer> tm = new TreeMap<>();
                for (int x : arr) tm.put(x, tm.getOrDefault(x, 0) + 1);
                long ans = fac[n];
                int max = tm.lastKey();
                int cnt1 = tm.pollLastEntry().getValue();
                Integer smax = tm.size() > 0 ? tm.lastKey() : null;
                if (smax == null) {
                    pw.println(ans);
                } else {
                    int smaxx = smax.intValue();
                    int cnt2 = tm.pollLastEntry().getValue();
                    if (smaxx != max - 1 && cnt1 == 1) {
                        pw.println(0);
                    } else {
                        int rem = n - cnt1 - cnt2;
                        int x = rem;
                        for (int i = cnt2 + cnt1; i < n; i++) {
//						pw.println(i+" "+rem+" "+x+" " +(fac[i-1]));
//						pw.println(ans+" "+((fac[rem]*nCrModPFermat(x,rem,mod))%mod*(fac[i-1]))%mod);
                            ans = sub(ans, (((((fac[rem] * cnt1) % mod) * nCrModPFermat(x, rem, mod)) % mod) * (fac[i - 1])) % mod, mod);
                            rem--;
                        }
                        ans = sub(ans, fac[n - 1], mod);
                        pw.println(ans);
                    }

                }
            }

        }

        long power(long x, long y, long p) {

            // Initialize result
            long res = 1;

            // Update x if it is more than or
            // equal to p
            x = x % p;

            while (y > 0) {

                // If y is odd, multiply x
                // with result
                if (y % 2 == 1)
                    res = (res * x) % p;

                // y must be even now
                y = y >> 1; // y = y/2
                x = (x * x) % p;
            }

            return res;
        }

        long modInverse(long n, long p) {
            return power(n, p - 2, p);
        }

        long nCrModPFermat(int n, int r,
                           long p) {

            if (n < r)
                return 0;
            // Base case
            if (r == 0)
                return 1;

            // Fill factorial array so that we
            // can find all factorial of r, n
            // and n-r
            return (fac[n] * modInverse(fac[r], p)
                    % p * modInverse(fac[n - r], p)
                    % p)
                    % p;
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

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

        public int[] nextIntArr(int n) {
            try {
                int[] a = new int[n];
                for (int i = 0; i < n; i++)
                    a[i] = nextInt();
                return a;
            } catch (Exception e) {
                throw new RuntimeException(e);
            }

        }

    }
}

