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 Div21569C {

	static long MOD = 998244353;
	static long[] factorial = new long[200001];
	
	public static void main(String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
			
		factorial[0] = 1;
		for (int i = 1; i <= 100000; i++) {
		    factorial[i] = factorial[i - 1] * i % MOD;
		}
		
		int t = Integer.parseInt(br.readLine());
		
		while(t --> 0) {
			
			int n = Integer.parseInt(br.readLine());
			StringTokenizer st = new StringTokenizer(br.readLine());
			
			ArrayList<Integer> ar = new ArrayList<Integer>();
			
			for(int i = 0; i < n; i++)
				ar.add(Integer.parseInt(st.nextToken()));
			
			Collections.sort(ar, Collections.reverseOrder());
			
			int mc = 1;
			int ouc = 0;
			int mbc = 0;
			int state = 0;
			
			for(int i = 1; i < n; i++) {
				
				if(ar.get(i) == ar.get(i-1)) {
					
					if(state == 0)
						mc++;
					else if(state == 1)
						ouc++;
					else
						mbc++;
					
				}else {
						
					if(ar.get(i) + 1 == ar.get(i-1))
						state++;
					else
						state += 2;
					
					if(state == 1)
						ouc++;
					else
						mbc++;
					
				}
				
			}
			
			if(mc >= 2 || n == 1) {
				
				System.out.println(factorial[n]);
				
			}else {
				
				if(ouc == 0) {
					
					System.out.println(0);
					
				}else {
					
					long val = factorial[n];
					
					//System.out.println(mbc + " " + ouc);
										
					for(int i = 0; i <= mbc; i++) {
						
						val = (val - binomial_coefficient(mbc, i) * factorial[ouc + mbc - i] * factorial[i]) % MOD;
						if(val < 0)
							val += MOD;
						//System.out.println(val + " " + i + " " + (ouc + mbc - i) + " " + binomial_coefficient(mbc, i));
						
					}
					
					System.out.println(val);
					
				}
				
			}
			
		}

	}
	
	static long binpow(long a, long b, long m) {
	    a %= m;
	    long res = 1;
	    while (b > 0) {
	        if ((b & 1) == 1)
	            res = res * a % m;
	        a = a * a % m;
	        b >>= 1;
	    }
	    return res;
	}
	
	static long binomial_coefficient(int n, int k) {
		if(k == 0)
			return 1;
	    return factorial[n] * binpow(factorial[k] * factorial[n - k] % MOD, MOD-2, MOD) % MOD;
	}

}
