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 C {
    static FastScanner sc;
    static PrintWriter pw;

    static int MOD = 998244353; 

    public static void main(String[] args) throws Exception{
        sc = new FastScanner();
        pw = new PrintWriter(System.out);
        
        try{

            int T = sc.ni();
            while(T-->0){
                int n = sc.ni();
                long a[] = new long[n];
                long max1 = Long.MIN_VALUE;
                long max2 = Long.MIN_VALUE;
                Map<Long,Integer> mp = new HashMap<>();
                for(int i=0; i<n; i++){
                    a[i] = sc.nl();
                    mp.put(a[i], mp.getOrDefault(a[i], 0)+1);
                    if(a[i]>max1){
                        max2 = max1;
                        max1 = a[i];
                    }else if(a[i]>max2){
                        max2 = a[i];
                    }
                }
                if(mp.get(max1)>=2){
                    long ans = factoial(n);
                    pw.println(ans);
                }else if(max2<=(max1-2)){
                    pw.println(0);
                }else{
                    int k = mp.get(max2);
                    long n1 = nCr(n, (k+1));
                    long n2 = 1;
                    long n3 = factoial(k);
                    long n4 = factoial(n-(k+1));
                    long n12 = ((n1%MOD)*(n2%MOD))%MOD;
                    long n34 = ((n3%MOD)*(n4%MOD))%MOD;
                    long invalid = ((n12%MOD)*(n34%MOD))%MOD;
                    long f1 = factoial(n);
                    long valid = ((f1%MOD)-(invalid%MOD))%MOD;
                    pw.println(valid); 
                }
            }

        }catch(Exception e){
            return;
        }

        pw.close();
    }

    static long factoial(int n){
        if (n >= MOD)
            return 0;
        long result = 1;
        for (int i = 1; i <= n; i++)
            result = ((result%MOD) * (i%MOD))%MOD;
     
        return result;
    }

    static long nCr(int n, int r){
		if (r > n - r)
			r = n - r;
		long C[] = new long[r + 1];
		C[0] = 1;
		for (int i = 1; i <= n; i++) {
			for (int j = Math.min(i, r); j > 0; j--)
				C[j] = ((C[j]%MOD) + (C[j - 1]%MOD)) % MOD;
		}
		return C[r];
	}

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in), 32768);
            st = null;
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int ni() {
            return Integer.parseInt(next());
        }

        int[] intArray(int N) {
            int[] ret = new int[N];
            for (int i = 0; i < N; i++)
                ret[i] = ni();
            return ret;
        }
 
        long nl() {
            return Long.parseLong(next());
        }

        long[] longArray(int N) {
            long[] ret = new long[N];
            for (int i = 0; i < N; i++)
                ret[i] = nl();
            return ret;
        }

        double nd() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}