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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class c {
    private static final int MAX = (int) 2e5 + 10;
    private static final int mod = 998244353;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        int tc = Integer.parseInt(br.readLine());

        long[] factorial = new long[MAX];
        factorial[0] = 1;
        for (int i = 1; i < MAX; i++) {
            factorial[i] = factorial[i - 1] * i;
            factorial[i] %= mod;
        }

        for (int tt = 0; tt < tc; tt++) {
            int n = Integer.parseInt(br.readLine());
            st = new StringTokenizer(br.readLine());
            List<Integer> list = new ArrayList<>();
            while (st.hasMoreTokens()) list.add(Integer.parseInt(st.nextToken()));

            list.sort(Integer::compareTo);
            int max = list.get(n - 1);
            int maxSecond = list.get(n - 2);

            int[] cnt = new int[2]; // 0 : max , 1 : secondMax
            for (int i = 0; i < n; i++) {
                if (list.get(i) == max) cnt[0]++;
                if (list.get(i) == maxSecond) cnt[1]++;
            }

//            최대값과 2번째 최대값이
//             0 차이,  최대값이 2개 이상 : 어케든 ㅇㅋ
//            1 차이 : 젤 큰거 뒤에 1차이 나는 거 둬야 함 ==젤 큰 거 뒤에 1차이 나는게 없는 상황 빼줘야 함
//            2 이상 차이 : 절대 안댐 ?

            long ans = factorial[n];

            if (max - maxSecond == 0) {
                ans = factorial[n];

            } else if (max - maxSecond == 1) {

                int othersCnt = n - 1 - cnt[1];

                for (int f = 0; f <= othersCnt; f++) {
                    long subtract = factorial[cnt[1] + f];

                    long A = factorial[othersCnt];
                    long B = (factorial[othersCnt - f] * factorial[f]) % mod;

                    long inverse = fastPow(B, mod - 2);
                    subtract *= ((A * inverse % mod) % mod);
                    subtract %= mod;
                    
                    subtract *= factorial[othersCnt - f];
                    subtract %= mod;

//                    System.out.println(B+" "+inverse+" "+(A*inverse));
//                    System.out.println(factorial[cnt[1]+f]+" "+((A * inverse) % mod)+" "+factorial[othersCnt - f]);
                    
                    ans+=mod;
                    ans -= subtract;
                    ans %= mod;
//                    System.out.println(subtract);
                }
            } else {
                ans = 0;
            }

            if (cnt[0] >= 2) {
                ans = factorial[n];
            }

            System.out.println(ans);
        }
    }


    private static long fastPow(long x, long e) {
        if (e == 0) return 1;
        if (e % 2 == 1) return (fastPow(x, e - 1) * x) % mod;
        else {
            long half = fastPow(x, e / 2) % mod;
            return (half * half) % mod;
        }
    }

}