#include <bits/stdc++.h>
using namespace std;
long long fact[200001];
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 998244353;
    }
    a = (a * a) % 998244353;
    b = b / 2;
  }
  return res;
}
long long solve(long long n, long long r) {
  if (r == 0) {
    return 1;
  }
  if (r > n) {
    return 1;
  }
  long long nfact = fact[n];
  long long nr = power(fact[n - r], 998244353 - 2);
  long long r1 = power(fact[r], 998244353 - 2);
  long long ans = nfact;
  ans = (ans * nr) % 998244353;
  ans = (ans * r1) % 998244353;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i <= 200000; i++) {
    fact[i] = (i * fact[i - 1]) % 998244353;
  }
  int tc = 1;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    if (a[n - 1] - a[n - 2] > 1) {
      cout << "0"
           << "\n";
      continue;
    }
    if (a[n - 1] == a[n - 2]) {
      cout << fact[n] << "\n";
    } else {
      long long temp = 0;
      long long ct = 0;
      int el = a[n - 2];
      int i = n - 2;
      while (i >= 0 && a[i] == el) {
        i--;
        ct++;
      }
      for (int k = n; k > ct; k--) {
        long long elb = k - 1;
        long long ela = n - k;
        if (elb == n - 1) {
          temp += fact[n - 1];
        } else {
          long long nel = n - 1 - ct;
          long long plb = elb - ct;
          long long k1 = solve(nel, plb);
          long long toad = (k1 * fact[elb]) % 998244353;
          toad = (toad * fact[ela]) % 998244353;
          temp = (temp + toad) % 998244353;
        }
      }
      cout << (fact[n] - temp + 998244353) % 998244353 << "\n";
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution 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;
long long fact[200001];
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 998244353;
    }
    a = (a * a) % 998244353;
    b = b / 2;
  }
  return res;
}
long long solve(long long n, long long r) {
  if (r == 0) {
    return 1;
  }
  if (r > n) {
    return 1;
  }
  long long nfact = fact[n];
  long long nr = power(fact[n - r], 998244353 - 2);
  long long r1 = power(fact[r], 998244353 - 2);
  long long ans = nfact;
  ans = (ans * nr) % 998244353;
  ans = (ans * r1) % 998244353;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i <= 200000; i++) {
    fact[i] = (i * fact[i - 1]) % 998244353;
  }
  int tc = 1;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    if (a[n - 1] - a[n - 2] > 1) {
      cout << "0"
           << "\n";
      continue;
    }
    if (a[n - 1] == a[n - 2]) {
      cout << fact[n] << "\n";
    } else {
      long long temp = 0;
      long long ct = 0;
      int el = a[n - 2];
      int i = n - 2;
      while (i >= 0 && a[i] == el) {
        i--;
        ct++;
      }
      for (int k = n; k > ct; k--) {
        long long elb = k - 1;
        long long ela = n - k;
        if (elb == n - 1) {
          temp += fact[n - 1];
        } else {
          long long nel = n - 1 - ct;
          long long plb = elb - ct;
          long long k1 = solve(nel, plb);
          long long toad = (k1 * fact[elb]) % 998244353;
          toad = (toad * fact[ela]) % 998244353;
          temp = (temp + toad) % 998244353;
        }
      }
      cout << (fact[n] - temp + 998244353) % 998244353 << "\n";
    }
  }
  return 0;
}

```