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.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long a[200005];
int main() {
  int t;
  scanf("%d", &t);
  for (int ti = 1; ti <= t; ti++) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    sort(a + 1, a + n + 1);
    if (a[n] == a[n - 1]) {
      long long res = 1;
      for (int i = 1; i <= n; i++) {
        res *= i;
        res %= mod;
      }
      printf("%lld\n", res);
    } else {
      if (a[n] == (a[n - 1] + 1)) {
        long long ans = 1;
        int res = 1;
        for (int j = (n - 1); j >= 1; j--) {
          if (a[j] == a[n - 1]) {
            res++;
          } else {
            break;
          }
        }
        for (int i = 1; i <= n; i++) {
          if (i == res) {
            ans *= (i - 1);
            ans %= mod;
            continue;
          }
          ans *= i;
          ans %= mod;
        }
        printf("%lld\n", ans);
      } else {
        printf("0\n");
      }
    }
  }
  return 0;
}
