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.InputStreamReader;
import java.util.StringTokenizer;

public class JuryMeeting{
	private static final int MOD = 998244353;
	public static void main(String args[]) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine().trim());
		var sb = new StringBuilder();
		while(t-- > 0){
			int n = Integer.parseInt(br.readLine().trim());
			int [] input = nextIntArray(n, br);
			sb.append(getValue(input)).append("\n");
		}
		br.close();
		System.out.println(sb);
	}

	private static long getValue(int [] input){
		int val = getMax(input);
		int countMax = getCount(input, val);
		int prevCount = getCount(input, val-1);
		
		long ret = 1, subtracter = 1;
		for(long i = 1; i  <= input.length; i++){
			ret = ret * i % MOD;
			if(i != prevCount+1) subtracter = subtracter * i % MOD;
		}
		
		if(countMax == 1) ret = (ret - subtracter + MOD) % MOD; 
		
		return ret;
	}

	private static int[] nextIntArray(int N, BufferedReader br) throws Exception {
		StringTokenizer st = new StringTokenizer(br.readLine().trim()," ");
		int arr[] = new int[N];
		for(int i = 0; i < N; i++)
		arr[i] = Integer.parseInt(st.nextToken());
		return arr;
	}
	
	private static int getCount(int [] input, int value) {
		int count = 0;
		for (int i : input)
			if (i == value)
				count++;

		return count;
	}

	private static int getMax(int ...arr) {
		int ret = Integer.MIN_VALUE;
		for(int i : arr)
			ret = Math.max(ret,i);
		return ret;
	}
}