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.
/*==========================================================================
*   	AUTHOR:    RonWonWon
*   	CREATED:   08.09.2021 23:41:43        
/*==========================================================================*/
import java.io.*;
import java.util.*;

public class C {
	
	public static void main(String[] args) throws IOException {
		/*
    in.ini();
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
    */
    long startTime = System.nanoTime();
    int t = in.nextInt(), T = 1;  
    while(t-->0) { 
      HashMap<Integer, Integer> freq = new HashMap<>();
      int n = in.nextInt();
      int a[] = in.readArray(n);
      for(int i : a) freq.put(i, freq.getOrDefault(i,0)+1);
      int m1 = 0, m2 = -1;
      for(int i : freq.keySet()) m1 = Math.max(m1, i);
      for(int i : freq.keySet()) if(m1!=i) m2 = Math.max(m2, i);
      if(freq.get(m1)>1){
      	long ans = 1;
      	for(int i=1;i<=n;i++) ans = (ans*i)%mod;
      	out.println(ans);
      }
      else{
      	if(m2==-1) out.println(0);
      	else if(m1-m2!=1) out.println(0);
      	else{
      		long ans = 0;
      		f = new long[n+1];
      		f[0] = 1; f[1] = 1;
      		for(int i=2;i<=n;i++) f[i] = (f[i-1] * i)%mod;
      		for(int i=0;i<=n-2;i++){
      			if(i<freq.get(m2)) ans = (ans + f[n-1])%mod;
      			else{
      				int diff = (i-freq.get(m2));
      				long ncr = nCk(n-freq.get(m2)-1,diff);
      				long minus = (f[i]*ncr)%mod;
      				minus = (minus*f[n-i-1])%mod;
      				ans = (ans + (f[n-1] - minus + mod)%mod)%mod;
      			}
      		}
      		out.println(ans);
      	}
      }  
   		//out.println("Case #"+T+": "+ans); T++;
    }
    /*
    startTime = System.nanoTime() - startTime;
    System.err.println("Time: "+(startTime/1000000));
    */
    out.flush();
	}
	
	static final long mod=998244353;
	static long f[];
	
	static long mul(long a, long b) {
		return a*b%mod;
	}

	static long fastPow(long base, long exp) {
		if (exp==0) return 1;
		long half=fastPow(base, exp/2);
		if (exp%2==0) return mul(half, half);
		return mul(half, mul(half, base));
	}
	
	static long modInv(long x) {
		return fastPow(x, mod-2);
	}
	
	static long fact(int x) {
		return f[x];
	}
	
	static long nCk(int n, int k) {
		return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n-k))));
	}
	
	static FastScanner in = new FastScanner();
  static PrintWriter out = new PrintWriter(System.out);
	static int oo = Integer.MAX_VALUE;
	static long ooo = Long.MAX_VALUE;
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		void ini() throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		long[] readArrayL(int n) {
			long a[] = new long[n];
			for(int i=0;i<n;i++) a[i] = nextLong();
			return a;
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
		
		double[] readArrayD(int n) {
			double a[] = new double[n];
			for(int i=0;i<n;i++) a[i] = nextDouble();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
	
	static void ruffleSortL(long[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n);
			long temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
}
