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.
// Compete Against yourself :)

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

public class TaskC {
    static FastReader fr = new FastReader();
    StringBuilder op = new StringBuilder();
    private final int mod = 998_244_353;

    public void run() {
        int tt=fr.nextInt();
        while(tt-->0){
            int n=fr.nextInt();
            int a[]=new int[n];
            for(int i=0;i<n;i++){
                a[i]=fr.nextInt();
            }
            int max=0;
            for(int x:a){
                max=Math.max(max,x);
            }
            int cnt=0;
            for(int x:a){
                if(x==max){
                    cnt++;
                }
            }
            int k=0;
            for(int x:a){
                if(x==max-1){
                    ++k;
                }
            }
            int ans=1,sub=1;
            for(int i=1;i<=n;i++){
                ans = mul(ans,i);
                if(i!=k+1){
                    sub = mul(sub,i);
                }
            }
            if(cnt==1){
                ans = (ans-sub+mod)%mod;
            }
            op.append(ans).append("\n");
        }
        System.out.println(op);
    }
    private int add(int a,int b){
        return (a%mod+b%mod)%mod;
    }
    private int mul(int a,int b){
        long ans=a;
        ans = (ans%mod);
        ans = (ans%mod)*(b%mod)%mod;
        return (int)(ans%mod);
    }

    public static void main(String[] args) {
        new TaskC().run();
    }
    //math related functions

    //gcd
    private long gcd(long a, long b) {
        if (a == 0) {
            return b;
        } else {
            return gcd(b % a, a);
        }
    }

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

    //exponential power
    private long power(long a, long b) {
        if (a == 1 || b == 0) {
            return 1;
        }
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                res = res * a;
            }
            b /= 2;
            a *= a;
        }
        return res;
    }

    // euler totient function copied from
    // https://www.geeksforgeeks.org/eulers-totient-function/

    private int phi(int n) {
        // Initialize result as n
        double result = n;

        // Consider all prime factors of n and for
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 2; p * p <= n; ++p) {
            // Check if p is a prime factor.
            if (n % p == 0) {
                // If yes, then update n and result
                while (n % p == 0)
                    n /= p;
                result *= (1.0 - (1.0 / (double) p));
            }
        }

        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (n > 1)
            result *= (1.0 - (1.0 / (double) n));

        return (int) result;
    }


    //binary search related functions
    private int lowerBound(int[] a, int key) {
        int ans = -1;
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = (low + high) >>> 1;
            if (a[mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    private int upperBound(int[] a, int key) {
        int ans = -1;
        int low = 0;
        int high = a.length - 1;
        while (low <= high) {
            int mid = (low + high) >>> 1;
            if (a[mid] > key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    //reverse
    private void reverse(int[] a, int s, int e) {
        for (int i = s, j = e; i <= j; i++, j--) {
            int temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }

    private void reverse(long[] a, int s, int e) {
        for (int i = s, j = e; i <= j; i++, j--) {
            long temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }

    //sorting
    private void sort(int[] a) {
        List<Integer> ls = new ArrayList<Integer>();
        for (int x : a) {
            ls.add(x);
        }
        Collections.sort(ls);
        for (int i = 0; i < a.length; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sort(long[] a) {
        List<Long> ls = new ArrayList<Long>();
        for (long x : a) {
            ls.add(x);
        }
        Collections.sort(ls);
        for (int i = 0; i < a.length; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sortInRangeInt(int[] a, int s, int e) {
        List<Integer> ls = new ArrayList<>();
        for (int i = s; i <= e; i++) ls.add(a[i]);
        Collections.sort(ls);
        for (int i = s; i <= e; i++) {
            a[i] = ls.get(i);
        }
    }

    private void sortInRangeLong(long[] a, int s, int e) {
        List<Long> ls = new ArrayList<>();
        for (int i = s; i <= e; i++) ls.add(a[i]);
        Collections.sort(ls);
        for (int i = s; i <= e; i++) {
            a[i] = ls.get(i);
        }
    }

    //Fast I/O
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        ;

        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }

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