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.util.*;
import java.io.*;

public class C_1569 {
	
	public static int pow(int a, int b, int m) {
		int res = 1;
		while(b > 0) {
			if((b & 1) == 1)
				res = (int)((1l * res * a) % m);
			a = (int)((1l * a * a) % m);
			b >>= 1;
		}
		return res;
	}
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int mod = 998244353;
		int[] fac = new int[(int)2e5 + 1];
		fac[0] = 1;
		for(int i = 1; i < fac.length; i++)
			fac[i] = (int)((1l * fac[i - 1] * i) % mod);
		
		int t = sc.nextInt();
		while(t-->0) {
			int n = sc.nextInt();
			int[] array = sc.nextIntArray(n);
			TreeMap<Integer, Integer> map = new TreeMap<>();
			for(int i = 0; i < n; i++)
				map.put(array[i], map.getOrDefault(array[i], 0) + 1);
			
			if(map.get(map.lastKey()) > 1) {
				pw.println(fac[n]);
			} else {
				if(map.lowerKey(map.lastKey()) != map.lastKey() - 1)
					pw.println(0);
				else {
					int b = map.pollLastEntry().getValue(), a = map.pollLastEntry().getValue();
					int all = fac[n];
					
					int ans = (int)((1l * fac[n]) % mod);
					int div = (int)((1l * fac[a + b] * fac[n - a - b]) % mod);
					div = pow(div, mod - 2, mod);
					ans = (int)((1l * ans * div) % mod);
					ans = (int)((1l * ans * fac[a + b]) % mod);
					
					int ans2 = (int)((1l * fac[n]) % mod);
					int div2 = (int)((1l * fac[a + b] * fac[n - a - b]) % mod);
					div2 = pow(div2, mod - 2, mod);
					ans2 = (int)((1l * ans2 * div2) % mod);
					ans2 = (int)((1l * ans2 * fac[a]) % mod);
					
					ans = (int)((1l * ans - ans2) % mod);
					ans = ((ans % mod) + mod) % mod;
					
					ans = (int)((1l * ans * fac[n - a - b]) % mod);
					
					//all = (int)((1l * all - ans) % mod);
					//all = ((all % mod) + mod) % mod;
					
					pw.println(ans);
				}
			}
			
		}
		
		pw.flush();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] array = new Integer[n];
			for (int i = 0; i < n; i++)
				array[i] = new Integer(nextInt());
			return array;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public static int[] shuffle(int[] a) {
			int n = a.length;
			Random rand = new Random();
			for (int i = 0; i < n; i++) {
				int tmpIdx = rand.nextInt(n);
				int tmp = a[i];
				a[i] = a[tmpIdx];
				a[tmpIdx] = tmp;
			}
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}
