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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.StringTokenizer;

public class taskc {
	long MOD = 998244353;

	public static void main(String[] args) {
		FastScanner in = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int tt = in.nextInt();
		for (int tc = 0; tc < tt; tc++) {
			taskc sol = new taskc();
			sol.solve(in, out);
		}
		out.flush();
	}

	void solve(FastScanner in, PrintWriter out) {
		int n = in.nextInt();
		long a[] = new long[n];
		for (int i = 0; i < n; i++) a[i] = in.nextLong();
		long fact[] = new long[n + 1];
		fact[0] = 1;
		for (int i = 1; i <= n; i++) {
			fact[i] = mul(i, fact[i - 1]);
		}
		HashMap<Long, Integer> m = new HashMap<>();
		long mx = 0;
		for (int i = 0; i < n; i++) {
			int cur = (m.containsKey(a[i]) ? m.get(a[i]) : 0);
			m.put(a[i], cur + 1);
			mx = Math.max(mx, a[i]);
		}
		int cnt = 0;
		for (int i = 0; i < n; i++) if (a[i] == mx) cnt++;

		if (cnt == n) {
			out.println(fact[n]);
		} else {
			if (m.containsKey(mx - 1)) {
				int cnt_second = 0;
				for (int i = 0; i < n; i++) {
					if (a[i] == mx - 1) {
						cnt_second++;
					}
				}
				long choose2From[] = new long[n+1];
				choose2From[cnt_second] = fact[cnt_second];
				long ans = choose2From[cnt_second];
				for (long i = cnt_second + 1; i < n; i++) {
					choose2From[(int) i] = mul(div(i, i - cnt_second), choose2From[(int) i - 1]);
					ans = add(ans, choose2From[(int) i]);
				}
				ans = mul(ans, fact[n - 1 - cnt_second]);
				ans = sub(fact[n], ans);
				out.println(ans);
			} else {
				out.println(0);
			}
		}
	}

	long add(long x, long y) {
		return ((x % MOD) + (y % MOD)) % MOD;
	}
	long sub(long x, long y) {
		return ((x % MOD) - (y % MOD)) % MOD;
	}
	long mul(long x, long y) {
		return ((x % MOD) * (y % MOD)) % MOD;
	}
	long exp(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 == 1) res = mul(res, x);
			x = mul(x, x);
			y /= 2;
		}
		return res;
	}
	long div(long x, long y) {
		return mul(x, exp(y, MOD - 2));
	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tok = new StringTokenizer("");
		String next() {
			while(!tok.hasMoreTokens()) {
				try {
					tok = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return tok.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		float nextFloat() {
			return Float.parseFloat(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}
