#include <bits/stdc++.h>
using namespace std;
int a[1111111];
int q;
int n;
int64_t gt[1111111];
int64_t Powe(int a, int n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  int64_t b = Powe(a, n / 2);
  b = (b * b) % 998244353;
  if (n % 2 == 1)
    return (b * a) % 998244353;
  else
    return b;
}
int main() {
  scanf("%d", &q);
  gt[0] = 1;
  for (int i = 1; i <= 1000000; i++) gt[i] = (gt[i - 1] * i) % 998244353;
  for (int z = 1; z <= q; z++) {
    scanf("%d", &n);
    int an, as = 0;
    bool ok = true;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (as < a[i]) {
        an = as;
        as = a[i];
      } else
        an = max(an, a[i]);
    }
    sort(a + 1, a + n + 1);
    int A = upper_bound(a + 1, a + n + 1, a[n]) -
            lower_bound(a + 1, a + n + 1, a[n]);
    if (A > 1) {
      cout << gt[n] << endl;
      ok = false;
    }
    if (a[n] - an > 1) {
      cout << 0 << endl;
      ok = false;
    }
    if (ok) {
      int B =
          upper_bound(a + 1, a + n + 1, an) - lower_bound(a + 1, a + n + 1, an);
      int64_t dd = (gt[B] * gt[n - B - 1]) % 998244353;
      int T = n - B - 1;
      int64_t X = 1;
      for (int i = B + 2; i <= n; i++) {
        int u = i - B - 1;
        int64_t cc = (gt[u] * gt[T - u]) % 998244353;
        cc = (gt[T] * Powe(cc, 998244353 - 2)) % 998244353;
        dd = (dd + (((gt[i - 1] * cc) % 998244353) * gt[n - i]) % 998244353) %
             998244353;
      }
      cout << (gt[n] - dd + 998244353) % 998244353 << endl;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[1111111];
int q;
int n;
int64_t gt[1111111];
int64_t Powe(int a, int n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  int64_t b = Powe(a, n / 2);
  b = (b * b) % 998244353;
  if (n % 2 == 1)
    return (b * a) % 998244353;
  else
    return b;
}
int main() {
  scanf("%d", &q);
  gt[0] = 1;
  for (int i = 1; i <= 1000000; i++) gt[i] = (gt[i - 1] * i) % 998244353;
  for (int z = 1; z <= q; z++) {
    scanf("%d", &n);
    int an, as = 0;
    bool ok = true;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (as < a[i]) {
        an = as;
        as = a[i];
      } else
        an = max(an, a[i]);
    }
    sort(a + 1, a + n + 1);
    int A = upper_bound(a + 1, a + n + 1, a[n]) -
            lower_bound(a + 1, a + n + 1, a[n]);
    if (A > 1) {
      cout << gt[n] << endl;
      ok = false;
    }
    if (a[n] - an > 1) {
      cout << 0 << endl;
      ok = false;
    }
    if (ok) {
      int B =
          upper_bound(a + 1, a + n + 1, an) - lower_bound(a + 1, a + n + 1, an);
      int64_t dd = (gt[B] * gt[n - B - 1]) % 998244353;
      int T = n - B - 1;
      int64_t X = 1;
      for (int i = B + 2; i <= n; i++) {
        int u = i - B - 1;
        int64_t cc = (gt[u] * gt[T - u]) % 998244353;
        cc = (gt[T] * Powe(cc, 998244353 - 2)) % 998244353;
        dd = (dd + (((gt[i - 1] * cc) % 998244353) * gt[n - i]) % 998244353) %
             998244353;
      }
      cout << (gt[n] - dd + 998244353) % 998244353 << endl;
    }
  }
}

```