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.*;

public class Main{
	static int mod=998244353;
	static long fastpow(long n,long ti) {
		 if (ti == 0) 
		        return 1l;
		if(ti%2==0) {
			long y=fastpow(n, ti/2);
			long k=y*y;
			k%=mod;
			return k;
		}
		else {
			long y=fastpow(n, ti/2);
			long k=((n*y)%mod)*y;
			k%=mod;
			return k;
		}
	}
	static long modInverse(long x) {
   	return fastpow(x, mod-2);
   }
	static long[]fac,facInv;
	static void preProcessNCR(int maxN) {
		//compute factorial for all numbers smaller than or equal maxN
		fac=new long[maxN+1];fac[0]=1;
		for(int i=1;i<fac.length;i++)fac[i]=(fac[i-1]*i)%mod;
		
		facInv=new long[fac.length];
		facInv[fac.length-1]=modInverse(fac[fac.length-1]);//modInverse(n!)
		for (int i=fac.length-2; i>=0; i--) {
			//modInverse(x!)=(x+1)/(x+1)!=(x+1)*modInverse((x+1)!)
			facInv[i]=(facInv[i+1]*(i+1))%mod;
		}
	}

	public static long ncr(int n,int r) {
		if(n<r)return 0;
		long num=fac[n];
		long den=(facInv[n-r]*facInv[r])%mod;
		long ans=(num*den)%mod;
		return ans;
	}
	static void main() throws Exception{
		int n=sc.nextInt();
		int[]in=sc.intSortedArr(n);
		for(int i=1;i<n;i++) {
			if(in[i]-in[i-1]>1) {
				pw.println(0);
				return;
			}
		}
		int cnt=0;
		for(int i=0;i<n;i++) {
			if(in[i]==in[n-1])cnt++;
		}
		if(cnt==1) {
			int cnt2=0;
			for(int i=0;i<n;i++) {
				if(in[i]==in[n-2]) {
					cnt2++;
				}
			}
			long ans=(ncr(n, cnt2+1)*fac[n-cnt2-1])%mod;
			long x=cnt2;
			ans=(ans*x)%mod;
			while(x>0) {
				ans=(ans*x)%mod;
				x--;
			}
			pw.println(ans);
			return;
		}
		pw.println(fac[n]);
	}
	public static void main(String[] args) throws Exception{
		sc=new MScanner(System.in);
//		sc=new MScanner("altstadt.in");
		pw = new PrintWriter(System.out);
		int tc=1;
        tc=sc.nextInt();
        preProcessNCR(300000);
        for(int i=1;i<fac.length;i++) {
        	fac[i]=(fac[i-1]*i)%mod;
        }
		for(curt=1;curt<=tc;curt++) {
//			pw.printf("Case #%d: ", i);
			main();
		}
		
		pw.flush();
	}
	static int curt;
	static PrintWriter pw;
	static MScanner sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
	 
		public MScanner(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 int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		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 boolean ready() throws IOException {
			return br.ready();
		}
	 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
		
	}
	static void dbg(int[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void dbg(long[]in) {
		System.out.println(Arrays.toString(in));
	}
	static void sort(int[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void sort(long[]in) {
		shuffle(in);
		Arrays.sort(in);
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}