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 BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
	static String readLine() throws IOException {
		return br.readLine();
	}
	static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(readLine());
		return st.nextToken();
	}
	static int readInt() throws IOException {
		return Integer.parseInt(next());
	}
	static long readLong() throws IOException {
		return Long.parseLong(next());
	}
	static double readDouble() throws IOException {
		return Double.parseDouble(next());
	}
	static char readChar() throws IOException {
		return next().charAt(0);
	}
	static class Pair implements Comparable<Pair> {
		int f, s;
		Pair(int f, int s) {
			this.f = f; this.s = s;
		}
		public int compareTo(Pair other) {
			if (this.f != other.f) return this.f - other.f;
			return this.s - other.s;
		}
	}
	final static int N = 200000, mod = 998244353;
	static long fac[] = new long[N + 1], inv[] = new long[N + 1];
	static long pow(long b, long p) {
		long ret = 1;
		while (p > 0) {
			if ((p&1) == 1) ret = b*ret%mod;
			b = b*b%mod;
			p >>= 1;
		}
		return ret;
	}
	static long C(int n, int k) {
		return fac[n]*(inv[k]*inv[n - k]%mod)%mod;
	}
	static void solve() throws IOException {
		int n = readInt(), max1 = 0, max2 = 0, a[] = new int[n + 1];
		for (int i = 1; i <= n; ++i) {
			a[i] = readInt();
			fac[i] = i*fac[i - 1]%mod;
			if (a[i] > max1) {
				max2 = max1;
				max1 = a[i];
			} else if (a[i] > max2) max2 = a[i];
		}
		if (max1 - max2 >= 2) {
			pr.println(0);
			return;
		}
		if (max1 - max2 == 0) {
			pr.println(fac[n]);
			return;
		}
		int cnt = 0;
		for (int i = 1; i <= n; ++i)
			if (a[i] >= max2) ++cnt;
		long bad = C(n, cnt);
		bad = bad*fac[cnt - 1]%mod;
		bad = bad*fac[n - cnt]%mod;
		pr.println((fac[n] - bad + mod)%mod);
	}
	public static void main(String[] args) throws IOException {
		//solve();
		fac[0] = inv[0] = 1;
		for (int i = 1; i <= N; ++i) {
			fac[i] = i*fac[i - 1]%mod;
			inv[i] = pow(fac[i], mod - 2);
		}
		for (int t = readInt(); t > 0; --t) solve();
		pr.close();
	}
}
