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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeMap;


public class Main {

	static class Task {
		
		int NN = 200005;
		int MOD = 998244353;
		int INF = 2000000000;
		long INFINITY = 2000000000000000000L;

		public void solve(InputReader in, PrintWriter out) {
			fac[0] = 1;
			for(int i=1;i<200005;++i) fac[i] = (fac[i - 1] * 1L * i)%MOD;
			int t = in.nextInt();
			while(t-->0){
				int n = in.nextInt();
				List<Integer> L = new ArrayList<>();
				for(int i=0;i<n;++i){
					L.add(in.nextInt());
				}
				Collections.sort(L);
				int d = L.get(n - 1) - L.get(n - 2);
				int c = 0;
				for(int i=n-2;i>=0 && L.get(i).equals(L.get(n - 2));--i)++c;
				if(d == 0) {
					out.println(fac[n]);
				} else if(d == 1) {
					out.println((fac[n] - (fac[n] * exp(c+1, MOD-2))%MOD + MOD)%MOD);
				} else {
					out.println(0);
				}
			}
		}

		long exp(long x, long n) {
			long ret = 1;
			long sq = x;
			while(n>0){
				if(n%2!=0){
					ret = (ret * sq) % MOD;
				}
				n /= 2;
				sq = (sq*sq)%MOD;
			}
			return ret;
		}

		long [] fac = new long[200005];

	}
	
	static void prepareIO(boolean isFileIO) {
		//long t1 = System.currentTimeMillis();
		Task solver = new Task();
		// Standard IO
		if(!isFileIO) { 
			InputStream inputStream = System.in;
	        OutputStream outputStream = System.out;
	        InputReader in = new InputReader(inputStream);
	        PrintWriter out = new PrintWriter(outputStream);
	        solver.solve(in, out);
	        //out.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        out.close();
		}
        // File IO
		else {
			String IPfilePath = System.getProperty("user.home") + "/Downloads/ip.in";
	        String OPfilePath = System.getProperty("user.home") + "/Downloads/op.out";
	        InputReader fin = new InputReader(IPfilePath);
	        PrintWriter fout = null;
	        try {
				fout = new PrintWriter(new File(OPfilePath));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
	        solver.solve(fin, fout);
	        //fout.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        fout.close();
		}
	}
	
	public static void main(String[] args) {
        prepareIO(false);
	}
	
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        
        public InputReader(String filePath) {
        	File file = new File(filePath);
            try {
				reader = new BufferedReader(new FileReader(file));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
            tokenizer = null;
        }
        
        public String nextLine() {
        	String str = "";
        	try {
				str = reader.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        	return str;
        }
        
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

}