#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const long long int qw = 998244353;
long long int power(long long int base, long long int exp, long long int M) {
  base %= M;
  long long int res = 1;
  while (exp > 0) {
    if (exp & 1) res = (res * base) % M;
    base = (base * base) % M;
    exp = exp >> 1;
  }
  return res;
}
long long int fac(long long int n, long long int M) {
  long long int ans = 1;
  if (n == 0 || n == 1) {
    return 1;
  }
  for (long long int i = 2; i < n + 1; i++) {
    ans = (ans * i) % M;
  }
  return ans;
}
long long int inverse(long long int n, long long int M) {
  long long int fa = fac(n, qw);
  return power(fa, M - 2, M);
}
long long int ncr(long long int n, long long int r, long long int M) {
  long long int num = fac(n, qw);
  long long int p1 = inverse(r, qw);
  long long int p2 = inverse(n - r, qw);
  long long int den = (p1 * p2) % qw;
  long long int ans = (num * den) % qw;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int arr[n];
    long long int k, m;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    priority_queue<long long int> pq(arr, arr + n);
    m = pq.top();
    pq.pop();
    if (pq.top() == m) {
      cout << fac(n, qw) << "\n";
    } else {
      k = pq.top();
      if (k != m - 1) {
        cout << "0\n";
        continue;
      }
      long long int count = 0;
      while (pq.empty() == false && pq.top() == k) {
        count++;
        pq.pop();
      }
      long long int nfac = fac(n, qw);
      long long int ans = fac(count, qw);
      ans = (ans * fac(n - (count + 1), qw)) % qw;
      ans = (ans * ncr(n, count + 1, qw)) % qw;
      long long int f = (((nfac - ans) % qw) + qw) % qw;
      cout << f << "\n";
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int MOD = 1000000007;
const long long int qw = 998244353;
long long int power(long long int base, long long int exp, long long int M) {
  base %= M;
  long long int res = 1;
  while (exp > 0) {
    if (exp & 1) res = (res * base) % M;
    base = (base * base) % M;
    exp = exp >> 1;
  }
  return res;
}
long long int fac(long long int n, long long int M) {
  long long int ans = 1;
  if (n == 0 || n == 1) {
    return 1;
  }
  for (long long int i = 2; i < n + 1; i++) {
    ans = (ans * i) % M;
  }
  return ans;
}
long long int inverse(long long int n, long long int M) {
  long long int fa = fac(n, qw);
  return power(fa, M - 2, M);
}
long long int ncr(long long int n, long long int r, long long int M) {
  long long int num = fac(n, qw);
  long long int p1 = inverse(r, qw);
  long long int p2 = inverse(n - r, qw);
  long long int den = (p1 * p2) % qw;
  long long int ans = (num * den) % qw;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int arr[n];
    long long int k, m;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    priority_queue<long long int> pq(arr, arr + n);
    m = pq.top();
    pq.pop();
    if (pq.top() == m) {
      cout << fac(n, qw) << "\n";
    } else {
      k = pq.top();
      if (k != m - 1) {
        cout << "0\n";
        continue;
      }
      long long int count = 0;
      while (pq.empty() == false && pq.top() == k) {
        count++;
        pq.pop();
      }
      long long int nfac = fac(n, qw);
      long long int ans = fac(count, qw);
      ans = (ans * fac(n - (count + 1), qw)) % qw;
      ans = (ans * ncr(n, count + 1, qw)) % qw;
      long long int f = (((nfac - ans) % qw) + qw) % qw;
      cout << f << "\n";
    }
  }
  return 0;
}

```