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.*;
import static java.lang.Math.*;

public class C{
	static long mod =  998244353;
	
	static long[] f, fi;
	
	static long pow(long b, long e) {
		if (e==0)return 1;
		long r = pow(b,e/2);
		r=r*r%mod;
		if (e%2==0) return r;
		return r*b%mod;
	}
	
	static long mi(long a) {
		return pow(a,mod-2);
	}
	
	
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		
		
		f = new long[(int)5e5];
		f[0] = 1;
		fi = new long[f.length];
		for (int i = 1; i < f.length; i++) {
			f[i] = f[i-1]*i%mod;
			fi[i] = mi(f[i]);
		}
		
		int t= readInt();
		test: while(t-->0) {
			int n =readInt();
			Long[] a = new Long[n];
			
			for (int i= 0; i < n; i++) a[i]=readLong();
			Arrays.sort(a);
			
			if (a[n-2] == a[n-1]) {
				out.println(f[n]);
				continue test;
			}
			
			if (a[n-1] - a[n-2] >=2) {
				out.println(0);
				continue;
			}
			
			int lo = 0;
			for (int i = 0; i< n; i++) {
				if (a[i] == a[n-2]) lo++;
			}
			
			//System.out.println("Testing " + n);
			
			
			long ans = f[lo];
			ans *=fi[lo+1];
			ans%=mod;
			ans = (1-ans+mod)%mod;
			out.println(f[n]*ans%mod);
		}
		
		out.close();
	}
	
	/* Stupid things to try if stuck:
	 * n=1, expand bsearch range
	 * brute force small patterns
	 * submit stupid intuition
	 */
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
}