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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.*;

public class C {
	private static final FastReader fs = new FastReader();
	private static final PrintWriter out = new PrintWriter(System.out);
	private static long m = 1_000_000_009;
	private static long p = 31;
	private static long mod = 998244353;
	private static long[] perm;
	
	private static void fill() {
		perm = new long[2_000_10];
		perm[0] = 1;
		for(int i = 1; i < 2_000_10; i++)
			perm[i] = (i * perm[i-1]) % mod;
	}
	
	private static void solve() {
		int n = fs.nextInt();
		int[]a = fs.readintarray(n);
		int max = Arrays.stream(a).max().getAsInt();
		long ans = perm[n];
		if(Arrays.stream(a).filter(ele -> ele > max-2).count() > 1) {
			if(Arrays.stream(a).filter(ele->ele == max).count() > 1)
				out.println(ans);
			else {
				long count = Arrays.stream(a).filter(e -> e <= max-2).count();
				ans = sub(ans, perm[n-1]);
				for(int i = 1; i <= count; i++) {
					ans = sub(ans, mul(mul(perm[(int) count], mPower(perm[(int) (count-i)], mod-2)), perm[n-i-1]));
				}
				out.println(ans);
			}
		}
		else
			out.println(0);
	}
 
	public static void main(String[] args) {
		Thread t = new Thread(null, null, "", 1 << 28) {
			public void run() {
				fill();
				int test_case = 1;
				test_case = fs.nextInt();
				for (int cs = 1; cs <= test_case; cs++)
					solve();
				out.close();
			}
		};
		t.start();
		try {
			t.join();
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}
 
	private static class FastReader {
		BufferedReader br;
		StringTokenizer st;
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
 
		int[] readintarray(int n) {
			int res[] = new int[n];
			for (int i = 0; i < n; i++)
				res[i] = nextInt();
			return res;
		}
 
		long[] readlongarray(int n) {
			long res[] = new long[n];
			for (int i = 0; i < n; i++)
				res[i] = nextLong();
			return res;
		}
	}
	
	private static int max(int... arg) {
		int length = arg.length;
		if(length == 0)
			throw null;
		int maxVal = arg[0];
		for(int i = 1; i < length; i++)
			maxVal = Math.max(maxVal, arg[i]);
		return maxVal;
	}
	
	private static int min(int... arg) {
		int length = arg.length;
		if(length == 0)
			throw null;
		int maxVal = arg[0];
		for(int i = 1; i < length; i++)
			maxVal = Math.min(maxVal, arg[i]);
		return maxVal;
	}
	
	private static long max(long... arg) {
		int length = arg.length;
		if(length == 0)
			throw null;
		long maxVal = arg[0];
		for(int i = 1; i < length; i++)
			maxVal = Math.max(maxVal, arg[i]);
		return maxVal;
	}
	
	private static long min(long... arg) {
		int length = arg.length;
		if(length == 0)
			throw null;
		long maxVal = arg[0];
		for(int i = 1; i < length; i++)
			maxVal = Math.min(maxVal, arg[i]);
		return maxVal;
	}
	
	private static int abs(int a) {
		return a >= 0 ? a : -a;
	}
	
	private static int abs(int a, int b) {
		return a-b >= 0 ? a-b : b-a;
	}
	
	private static long abs(long a) {
		return a >= 0 ? a : -a;
	}
	
	private static long abs(long a, long b) {
		return a-b >= 0 ? a-b : b-a;
	}
	
	private static int power(int base, int exponent) {
		int res = 1;
		while(exponent > 0) {
			if((exponent & 1) != 0)
				res *= base;
			exponent >>= 1;
			base *= base;
		}
		return res;
	}
	
	private static long power(long base, long exponent) {
		long res = 1;
		while(exponent > 0) {
			if((exponent & 1) != 0)
				res *= base;
			exponent >>= 1;
			base *= base;
		}
		return res;
	}
	
	private static long mPower(long base, long exponent) {
		long res = 1;
		base %= mod;
		while(exponent > 0) {
			if((exponent & 1) != 0)
				res = mul(res, base);
			exponent >>= 1;
			base = mul(base, base);
		}
		return res;
	}
	
	private static long mul(long a, long b) {
		return a*b % mod;
	}
	
	private static long add(long a, long b) {
		return (a+b) % mod;
	}
	
	private static long sub(long a, long b) {
		long res = (a - b) % mod;
		if(res < 0)
			res += mod;
		return res;
	}
}