#include <bits/stdc++.h>
using namespace std;
void _print(vector<long long> v) {
  cout << "[ ";
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "]"
       << "\n";
}
long long Power(long long $a, long long $b) {
  long long $result = 1;
  $a %= 998244353ll;
  while ($b) {
    if ($b % 2 == 1) $result *= $a;
    $a *= $a;
    $a %= 998244353ll;
    $result %= 998244353ll;
    $b /= 2;
  }
  return $result;
}
long long Mod_Inv(long long $a) {
  $a %= 998244353ll;
  $a = Power($a, 998244353ll - 2);
  return $a;
}
long long Factorial[500001];
long long Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < 500001; i++) {
    Factorial[i] = i * Factorial[i - 1];
    Factorial[i] %= 998244353ll;
  }
  return 0;
}
long long Implement_Make_Factorial = Make_Factorial();
long long nCr(long long $n, long long $r) {
  if ($n < $r) return 0;
  long long $ans = Factorial[$n];
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$r]);
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$n - $r]);
  $ans %= 998244353ll;
  return $ans;
}
long long nPr(long long $n, long long $r) {
  long long $ans = Factorial[$n];
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$n - $r]);
  $ans %= 998244353ll;
  return $ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  set<long long> s;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    s.insert(arr[i]);
    if (mp.find(arr[i]) == mp.end())
      mp[arr[i]] = 1;
    else
      mp[arr[i]]++;
  }
  sort(arr.begin(), arr.end());
  if (arr[n - 1] - arr[n - 2] >= 2 && mp[arr[n - 1]] == 1) {
    cout << 0 << "\n";
    return;
  }
  long long ans = Factorial[n];
  if (mp[arr[n - 1]] > 1) {
    cout << ans << "\n";
    return;
  }
  long long cnt = mp[arr[n - 2]] + 1;
  ans *= Mod_Inv(Factorial[cnt]);
  ans %= 998244353ll;
  ans *= Factorial[cnt - 1];
  ans %= 998244353ll;
  long long anss =
      (Factorial[n] % 998244353ll - ans % 998244353ll) % 998244353ll;
  if (anss < 0) anss += 998244353ll;
  cout << anss << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void _print(vector<long long> v) {
  cout << "[ ";
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "]"
       << "\n";
}
long long Power(long long $a, long long $b) {
  long long $result = 1;
  $a %= 998244353ll;
  while ($b) {
    if ($b % 2 == 1) $result *= $a;
    $a *= $a;
    $a %= 998244353ll;
    $result %= 998244353ll;
    $b /= 2;
  }
  return $result;
}
long long Mod_Inv(long long $a) {
  $a %= 998244353ll;
  $a = Power($a, 998244353ll - 2);
  return $a;
}
long long Factorial[500001];
long long Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < 500001; i++) {
    Factorial[i] = i * Factorial[i - 1];
    Factorial[i] %= 998244353ll;
  }
  return 0;
}
long long Implement_Make_Factorial = Make_Factorial();
long long nCr(long long $n, long long $r) {
  if ($n < $r) return 0;
  long long $ans = Factorial[$n];
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$r]);
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$n - $r]);
  $ans %= 998244353ll;
  return $ans;
}
long long nPr(long long $n, long long $r) {
  long long $ans = Factorial[$n];
  $ans %= 998244353ll;
  $ans *= Mod_Inv(Factorial[$n - $r]);
  $ans %= 998244353ll;
  return $ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  set<long long> s;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    s.insert(arr[i]);
    if (mp.find(arr[i]) == mp.end())
      mp[arr[i]] = 1;
    else
      mp[arr[i]]++;
  }
  sort(arr.begin(), arr.end());
  if (arr[n - 1] - arr[n - 2] >= 2 && mp[arr[n - 1]] == 1) {
    cout << 0 << "\n";
    return;
  }
  long long ans = Factorial[n];
  if (mp[arr[n - 1]] > 1) {
    cout << ans << "\n";
    return;
  }
  long long cnt = mp[arr[n - 2]] + 1;
  ans *= Mod_Inv(Factorial[cnt]);
  ans %= 998244353ll;
  ans *= Factorial[cnt - 1];
  ans %= 998244353ll;
  long long anss =
      (Factorial[n] % 998244353ll - ans % 998244353ll) % 998244353ll;
  if (anss < 0) anss += 998244353ll;
  cout << anss << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```