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.
/*Everything is Hard 
 * Before Easy 
 * Jai Mata Dii 
 */ 
 
import java.util.*;
import java.io.*; 
  
public class Main {
	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; }} 
//	static long mod = (long)(1e9+7); 
	 static long mod = 998244353; 
//	 static Scanner sc = new Scanner(System.in); 
	static FastReader sc = new FastReader(); 
	static PrintWriter out = new PrintWriter(System.out);
	public static void main (String[] args) { 
		int ttt = 1;
		ttt = sc.nextInt();
		precompute();
		z :for(int tc=1;tc<=ttt;tc++){
			int n = sc.nextInt();
			int a[] = new int[n];
			for(int i=0;i<n;i++) {
				a[i] = sc.nextInt();
			}
			sort(a);
			int max = 0, cnt = 0, smax = 0;
			for(int i=0;i<n;i++) {
				if(a[i]>max) {
					smax = max;
					max = a[i];
					cnt = 1;
				}
				else if(a[i] == max) {
					cnt++;
				}
				else if(smax > a[i]) {
					smax = a[i];
				}
			}
			if(cnt == 1 && (max - smax)>1) {
				out.write("0\n");
				continue;
			}
			long ans = 1;
			for(long i = 1;i<=n;i++) {
				ans = (ans * i)%mod;
			}
			if(cnt == 1) {
				int less = 0;
				for(int i=0;i<n;i++) {
					if(a[i]<max-1) less++;
				}
				for(int i=0;i<=less;i++) {
					long mul = ((fac[(int)less] * power(fac[(int)(less-i)],mod-2,mod))%mod * power(fac[(int)i],mod-2,mod)%mod)%mod;
//					System.out.println(i+" "+mul);
					long toMinus = 1L*mul*(fac[i]*fac[n-i-1]%mod)%mod;
					ans = (ans - toMinus + mod)%mod;
				}
			}
			out.write(ans+"\n");
		}
		out.close();
	}
	static long fac[] = new long[300006];
	
	static void precompute() {
		int n = 300005;
		fac[0] = 1;
		for(int i=1;i<=n;i++)
			fac[i] = (i * fac[i-1])%mod;
	}
	
	static long power(long base, long p, long mod2) {
		
		long ans = 1;
		while(p!=0) {
			if(p%2==0) {
				base = (base*base)%mod2;
				p/=2;
			}
			else {
				ans = (ans*base)%mod2;
				p-=1;
			}
		}
		return (ans%mod2);
	}
	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
}