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.
//Utilities
import java.io.*;
import java.util.*;

public class C {
	static int t;
	static int n;
	static long[] a;
	static long[] fac;
	
	static final int MOD = 998244353;
	
	public static void main(String[] args) throws IOException {
		fac = new long[200005];
		fac[0] = 1;
		for (int i = 1; i < 200005; i++) {
			fac[i] = fac[i-1] * i % MOD;
		}
		t = in.iscan();
		while (t-- > 0) {
			run();
		}
		out.close();
	} 
	
	static void run() throws IOException {
		n = in.iscan(); a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.lscan();
		}
		UTILITIES.sort(a, true);
		long cntMax = 0, cntNextMax = 0;
		long max = a[n-1], nextMax = 0;
		for (int i = n-1; i >= 0; i--) {
			if (a[i] == a[n-1]) {
				cntMax++;
			}
			else {
				nextMax = a[i];
				break;
			}
		}
		for (int i = n-1; i >= 0; i--) {
			if (a[i] == nextMax) {
				cntNextMax++;
			}
		}
		if (cntMax == 1) {
			if (max - nextMax > 1) {
				out.println(0);
				return;
			}
			long res = 0;
			for (int i = n-1; i >= 1; i--) {
				res = (res + (long)(n-i) * fac[(int)(cntNextMax)] % MOD * fac[(int)(n-cntNextMax-1)] % MOD) % MOD;
			}
			res = res * (cntNextMax) % MOD;
			out.println(res);
		}
		else { // cntMax > 1
			out.println(fac[n]);
		}
	}
	
	static long choose(int n, int k) {
		if (k > n) {
			return 0;
		}
		long a = fac[n];
		long b = fac[k] * fac[n-k] % MOD;
		return a * UTILITIES.fast_pow_mod(b, MOD-2, MOD) % MOD;
	}
	
	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;
		
		public static void sort(int[] a, boolean increasing) {
			ArrayList<Integer> arr = new ArrayList<Integer>();
			int n = a.length;
			for (int i = 0; i < n; i++) {
				arr.add(a[i]);
			}
			Collections.sort(arr);
			for (int i = 0; i < n; i++) {
				if (increasing) {
					a[i] = arr.get(i);
				}
				else {
					a[i] = arr.get(n-1-i);
				}
			}
		}
		
		public static void sort(long[] a, boolean increasing) {
			ArrayList<Long> arr = new ArrayList<Long>();
			int n = a.length;
			for (int i = 0; i < n; i++) {
				arr.add(a[i]);
			}
			Collections.sort(arr);
			for (int i = 0; i < n; i++) {
				if (increasing) {
					a[i] = arr.get(i);
				}
				else {
					a[i] = arr.get(n-1-i);
				}
			}
		}
		
		public static void sort(double[] a, boolean increasing) {
			ArrayList<Double> arr = new ArrayList<Double>();
			int n = a.length;
			for (int i = 0; i < n; i++) {
				arr.add(a[i]);
			}
			Collections.sort(arr);
			for (int i = 0; i < n; i++) {
				if (increasing) {
					a[i] = arr.get(i);
				}
				else {
					a[i] = arr.get(n-1-i);
				}
			}
		}

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}
		
		public static void updateMap(HashMap<Integer, Integer> map, int key, int v) {
			if (!map.containsKey(key)) {
				map.put(key, v);
			}
			else {
				map.put(key, map.get(key) + v);
			}
			if (map.get(key) == 0) {
				map.remove(key);
			}
		}

		public static long gcd (long a, long b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static long lcm (long a, long b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
		
		// start of permutation and lower/upper bound template
		public static void nextPermutation(int[] nums) {
		    //find first decreasing digit
		    int mark = -1;
		    for (int i = nums.length - 1; i > 0; i--) {
		        if (nums[i] > nums[i - 1]) {
		            mark = i - 1;
		            break;
		        }
		    }
		 
		    if (mark == -1) {
		        reverse(nums, 0, nums.length - 1);
		        return;
		    }
		 
		    int idx = nums.length-1;
		    for (int i = nums.length-1; i >= mark+1; i--) {
		        if (nums[i] > nums[mark]) {
		            idx = i;
		            break;
		        }
		    }
		 
		    swap(nums, mark, idx);
		 
		    reverse(nums, mark + 1, nums.length - 1);
		}
		 
		public static void swap(int[] nums, int i, int j) {
		    int t = nums[i];
		    nums[i] = nums[j];
		    nums[j] = t;
		}
		 
		public static void reverse(int[] nums, int i, int j) {
		    while (i < j) {
		        swap(nums, i, j);
		        i++;
		        j--;
		    }
		}
		
		static int lower_bound (int[] arr, int hi, int cmp) {
			int low = 0, high = hi, mid = -1;
			while (low < high) {
				mid = (low + high) / 2;
				if (arr[mid] >= cmp) high = mid;
				else low = mid + 1;
			}
			return low;
		}
	 
		static int upper_bound (int[] arr, int hi, int cmp) {
			int low = 0, high = hi, mid = -1;
			while (low < high) {
				mid = (low + high) / 2;
				if (arr[mid] > cmp) high = mid;
				else low = mid + 1;
			}
			return low;
		}
		// end of permutation and lower/upper bound template
	}
}
