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 eduprac implements Runnable 
{
	private boolean console=false;

	long f[]=new long[1000000];
	long m=998244353;
	public void solve() 
	{
		int i; 	int n=in.ni(); Integer a[]=new Integer[n]; 
		for(i=0;i<n;i++) a[i]=in.ni();
		Arrays.sort(a);
		int b=a[n-1]; int c1=a[n-2];
		if(b==c1)
		{
			out.println(f[n]); return;
		}
		else
		{
			if(a[n-1]-a[n-2]!=1)
			{
				out.println(0); return;
			}
			int c=0;
			for(i=0;i<n;i++)
			{
				int v=a[i]; int v2=a[n-2];
				if(v==v2) c++;
			}
			int cf=n-c-1;
			int start=n-1; long cnt=0;
			while(start>=c)
			{
				long cur = nCr(start,c,m);
				cnt = (cnt+cur)%m;
				start--;
			}
			cnt = (cnt*f[cf])%m;
			cnt = (cnt*f[c])%m;
			long ans= (f[n]-cnt+m)%m;
			out.println(ans);
		}
	}
	public long power(long a,long b,long mod)
	{
		long x = 1, y = a; 
		while (b > 0)
		{
			if (b%2!=0) 
				x = (x*y)%mod;
			y = (y*y)%mod; 
			b /= 2; 
		} 
		return x%mod; 
	}
	long modinverse(long n, long mod)
	{ 
		return power(n, mod-2, mod); 
	}
	long nCr(int n, int r, long mod)
	{ 
		return (f[n]*((modinverse(f[r], mod) * modinverse(f[n-r], mod)) % mod)) % mod; 
	}
	@Override
	public void run() {
		try {  init();  } 
		catch (FileNotFoundException e) {  e.printStackTrace();   }

		f[0]=1; int i;
		for(i=1;i<f.length;i++)
		{
			f[i]=i*f[i-1];
			f[i]%=m;
		}
		int t= in.ni();

		while(t-->0)
		{
			solve();
			out.flush(); 
		}
	}
	private FastInput in;    private PrintWriter out;
	public static void main(String[] args) throws Exception {	 new eduprac().run();	 }

	private void init() throws FileNotFoundException {
		InputStream inputStream = System.in; 	 OutputStream outputStream = System.out;
		try {  if (!console && System.getProperty("user.name").equals("sachan")) {
			outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
			inputStream = new FileInputStream("/home/sachan/Desktop/input.txt");   	}
		}	catch (Exception ignored) {	}
		out = new PrintWriter(outputStream); 	 in = new FastInput(inputStream);
	}
	static class FastInput { InputStream obj;
	public FastInput(InputStream obj) {	this.obj = obj;	}
	byte inbuffer[] = new byte[1024];  int lenbuffer = 0, ptrbuffer = 0;
	int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
	if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
	try { lenbuffer = obj.read(inbuffer);  }
	catch (IOException e) { throw new InputMismatchException(); } }    
	if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }

	String ns() { int b = skip();StringBuilder sb = new StringBuilder();
	while (!(isSpaceChar(b)))	  { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}

	int ni() { int num = 0, b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); }}

	long nl() { long num = 0;int b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); } }

	boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
	int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
	float nf() {return Float.parseFloat(ns());}
	double nd() {return Double.parseDouble(ns());}
	char nc() {return (char) skip();}
	}
}