#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  for (int ii = 0; ii < t; ii++) {
    long long n;
    cin >> n;
    long long A[200000];
    for (long long i = 0; i < n; i++) {
      cin >> A[i];
    }
    long long mmax = A[0];
    for (long long i = 1; i < n; i++) {
      mmax = max(mmax, A[i]);
    }
    long long b = 0;
    for (long long i = 0; i < n; i++) {
      if (A[i] == mmax) b++;
    }
    if (b > 1) {
      long long c = 1;
      for (long long i = 2; i <= n; i++) {
        c *= i;
        c %= 998244353;
      }
      cout << c << endl;
      continue;
    }
    b = 0;
    for (long long i = 0; i < n; i++) {
      if (A[i] == mmax - 1) b++;
    }
    if (b > 0) {
      {
        n--;
        int k = n - b;
        long long N[200000], B[200000];
        long long c = 1;
        for (long long i = 1; i <= n + 1; i++) {
          c *= i;
          c %= 998244353;
          N[i] = c;
        }
        c = 1;
        for (long long i = k; i >= 1; i--) {
          c *= i;
          c %= 998244353;
          B[i] = c;
        }
        long long s = N[n];
        long long h;
        for (long long i = 0; i < k; i++) {
          h = (B[k - i] * N[n - 1 - i]) % 998244353;
          s += h;
          s %= 998244353;
        }
        long long ff = N[n + 1] - s;
        if (ff <= 0) ff += 998244353;
        cout << ff << endl;
      }
      continue;
    } else {
      cout << 0 << endl;
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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 main() {
  int t;
  cin >> t;
  for (int ii = 0; ii < t; ii++) {
    long long n;
    cin >> n;
    long long A[200000];
    for (long long i = 0; i < n; i++) {
      cin >> A[i];
    }
    long long mmax = A[0];
    for (long long i = 1; i < n; i++) {
      mmax = max(mmax, A[i]);
    }
    long long b = 0;
    for (long long i = 0; i < n; i++) {
      if (A[i] == mmax) b++;
    }
    if (b > 1) {
      long long c = 1;
      for (long long i = 2; i <= n; i++) {
        c *= i;
        c %= 998244353;
      }
      cout << c << endl;
      continue;
    }
    b = 0;
    for (long long i = 0; i < n; i++) {
      if (A[i] == mmax - 1) b++;
    }
    if (b > 0) {
      {
        n--;
        int k = n - b;
        long long N[200000], B[200000];
        long long c = 1;
        for (long long i = 1; i <= n + 1; i++) {
          c *= i;
          c %= 998244353;
          N[i] = c;
        }
        c = 1;
        for (long long i = k; i >= 1; i--) {
          c *= i;
          c %= 998244353;
          B[i] = c;
        }
        long long s = N[n];
        long long h;
        for (long long i = 0; i < k; i++) {
          h = (B[k - i] * N[n - 1 - i]) % 998244353;
          s += h;
          s %= 998244353;
        }
        long long ff = N[n + 1] - s;
        if (ff <= 0) ff += 998244353;
        cout << ff << endl;
      }
      continue;
    } else {
      cout << 0 << endl;
    }
  }
  return 0;
}

```