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.
//This code is written by प्रविण शंखपाळ 

//package wizard;

import java.io.*;
import java.math.*;
import java.util.*;

import java.text.*;

public class Ginny_Weasley {

	static long mod = (long) 998244353;

	public static void main(String[] args) {

//		try {

		FastReader fr = new FastReader();
		PrintWriter pt = new PrintWriter(System.out);

		int t = fr.nextInt();

		while (t > 0) {

			int n = fr.nextInt();

			long A[] = new long[n];
			for (int i = 0; i < n; i++) {
				A[i] = fr.nextLong();
			}

			sort(A, 0, n - 1);

			long fac = 1, ans = 0, k = 0, fac1 = 1;

			for (long i = 1; i <= (long) n; i++) {
				fac *= i;
				fac %= mod;
			}

			for (int i = 0; i < n - 1; i++) {
				if (A[i] == A[n - 1] - 1) {
					k++;
				}
			}

			for (long i = 1; i <= (long) n; i++) {
				if (i != k + 1) {
					fac1 *= i;
					fac1 %= mod;
				}
			}

			if (A[n - 1] == A[n - 2]) {
				ans = fac;
			} else if (A[n - 1] - A[n - 2] >= 2) {
				ans = 0;
			} else {

				ans = (((fac1 * k) % mod));
				ans %= mod;

			}

			pt.println(ans);

			t--;
		}
		pt.close();

//		} catch (
//
//		Exception e) {
//			return;
//		}
	}

	static int lower_bound(long[] arr, int begin, int end, long tar) {
		while (begin < end) {
			int mid = begin + (end - begin) / 2;
			if (arr[mid] < tar)
				begin = mid + 1;
			else if (arr[mid] >= tar)
				end = mid;
		}
		return begin;
//		lower bound gives largest index of a number smaller than (if target is not present) or equal to target(if target is present) 
//      for complete array use begin = 0 and end = n
	}

	static int upper_bound(long[] arr, int begin, int end, long tar) {
		while (begin < end) {
			int mid = begin + (end - begin) / 2;
			if (arr[mid] <= tar)
				begin = mid + 1;
			else if (arr[mid] > tar)
				end = mid;
		}

		return begin;
//		upper bound gives smallest index of a number strictly greater than target
//	    for complete array use begin = 0 and end = n
	}

	public static void mergeSort(String[] nameGo) {
		if (nameGo.length > 1) {
			String[] leftGo = new String[nameGo.length / 2];
			String[] rightGo = new String[nameGo.length - nameGo.length / 2];
			for (int so = 0; so < leftGo.length; so++) {
				leftGo[so] = nameGo[so];
			}
			for (int ki = 0; ki < rightGo.length; ki++) {
				rightGo[ki] = nameGo[ki + nameGo.length / 2];
			}
			mergeSort(leftGo);
			mergeSort(rightGo);
			merges(nameGo, leftGo, rightGo);
		}
	}

	public static void merges(String[] nameH, String[] leftH, String[] rightH) {
		int as = 0;
		int bs = 0;
		for (int i = 0; i < nameH.length; i++) {
			if (bs >= rightH.length || (as < leftH.length && leftH[as].compareToIgnoreCase(rightH[bs]) < 0)) {
				nameH[i] = leftH[as];
				as++;
			} else {
				nameH[i] = rightH[bs];
				bs++;
			}
		}
	}

	static void merge(long arr[], int l, int m, int r) {

		int n1 = m - l + 1;
		int n2 = r - m;

		long L[] = new long[n1];
		long R[] = new long[n2];

		for (int i = 0; i < n1; ++i)
			L[i] = arr[l + i];
		for (int j = 0; j < n2; ++j)
			R[j] = arr[m + 1 + j];

		int i = 0, j = 0;

		int k = l;
		while (i < n1 && j < n2) {
			if (L[i] <= R[j]) {
				arr[k] = L[i];
				i++;
			} else {
				arr[k] = R[j];
				j++;
			}
			k++;
		}

		while (i < n1) {
			arr[k] = L[i];
			i++;
			k++;
		}

		while (j < n2) {
			arr[k] = R[j];
			j++;
			k++;
		}
	}

	static void sort(long arr[], int l, int r) {
		if (l < r) {

			int m = l + (r - l) / 2;

			sort(arr, l, m);
			sort(arr, m + 1, r);

			merge(arr, l, m, r);
		}
	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			if (this.a != o.a)
				return Integer.compare(this.a, o.a);
			else
				return Integer.compare(this.b, o.b);
			// return 0;
		}

		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.a == a && p.b == b;
			}
			return false;
		}

	}

	static int binarySearch(int arr[], int first, int last, int key) {
		int mid = (first + last) / 2;
		while (first <= last) {
			if (arr[mid] < key) {
				first = mid + 1;
			} else if (arr[mid] == key) {
				return mid;
			} else {
				last = mid - 1;
			}
			mid = (first + last) / 2;
		}
		return -1;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
