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

public class B{
    static void solve(String a, int cnt1, char[][] ans){
        int first = -1, second = -1;
        int initial = -1;
        int cnt = 0;
        int n = a.length();
        for(int i=0;i<n;i++){
            if(a.charAt(i) == '1')
                continue;
            if(first == -1){
                first = i;
                initial = i;
                cnt++;
                continue;
            }
            second = i;
//            System.out.println("first: "+first+" second: "+second+" initial "+initial + "cnt1: "+cnt1+" cnt: "+cnt);
            ans[first][second] = '+';
            ans[second][first] = '-';
            first = second;
//            second = -1;
            cnt++;
            if(cnt == cnt1) {
                ans[first][initial] = '+';
                ans[initial][first] = '-';
//                return;
            }
        }
    }
    static int MOD = 998244353;
    static int[] fact;
    static void getfact(){
        fact[0] = 1;
        for(int i=1;i<=200000;i++){
            fact[i] = (fact[i-1]*i)%MOD;
        }
    }
    static int gcdExtended(int a,int b, Integer x, Integer y){
        if(a == 0)
        {
            x = 0;
            y = 1;
            return b;
        }
        Integer x1=1, y1=1;
        int gcd = gcdExtended(b%a, a, x1, y1);
        x = y1 - (b/a)*x1;
        y = x1;
        return gcd;
    }
    static int modInverse(int a){
            int m0 = MOD, m = MOD;
            int y = 0,x = 1;
            if (m == 1)
                return 0;

            while (a > 1) {
                int q = a / m;
                int t = m;
                m = a % m;
                a = t;
                t = y;
                y = x - q * y;
                x = t;
            }
            if (x < 0)
                x += m0;

            return x;
    }
    static int modDivide(int a, int b){
        a = a%MOD;
        int inv = modInverse(b);
        int c = (inv*a)%MOD;
        return c;
    }
    static int nCr(int n,int r){
        int sum = fact[n];
        sum = modDivide(sum,fact[r]);
        sum = modDivide(sum,fact[n-r]);
        return sum;
    }
    public static void main(String[] args){
        FastScanner fs = new FastScanner();
        int t = fs.nextInt();
        fact = new int[200001];
        getfact();
        while(t>0){
            int n = fs.nextInt();
            int[] arr = fs.readArray(n);
            int largest = 0, second_largest = 0, freq_largest = 0, freq_second_largest = 0;
            for(int i=0;i<n;i++){
                if(arr[i]>largest) {
                    second_largest = largest;
                    freq_second_largest = 1;
                    largest = arr[i];
                    freq_largest = 1;
                    continue;
                }else if(arr[i] == largest){
                    freq_largest++;
                }else if(arr[i] == second_largest)
                    freq_second_largest++;
            }
            if(freq_largest > 1)
                System.out.println(fact[n]);
            else if(largest - second_largest > 1)
                System.out.println(0);
            else{
                int curr = fact[n];
                int cnt = n - 1 - freq_second_largest;
                for(int i=0;i<=cnt;i++){
                    int sum = (nCr(cnt,i)*(fact[i]*fact[n-i-1])%MOD)%MOD;
                    curr = curr - sum;
                }
                System.out.println(curr);
            }
            t--;
        }
        return;
    }
    static class FastScanner{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next(){
            while(!st.hasMoreTokens()){
                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());
        }
        int[] readArray(int n){
            int[] a = new int[n];
            for(int i=0;i<n;i++)a[i] = nextInt();
            return a;
        }
    }
}