#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int b, long long int p = 1) {
  x %= 998244353;
  while (b > 0) {
    if (b & 1) p = (p * x) % 998244353;
    b >>= 1;
    x = (x * x) % 998244353;
  }
  return p;
}
vector<long long int> fact;
void fact_pre(long long int n = 1100000) {
  fact.resize(n, 1);
  for (long long int i = 2; i < n; i++) fact[i] = (fact[i - 1] * i) % 998244353;
}
long long int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  return (fact[n] * power((fact[n - r] * fact[r]) % 998244353, 998244353 - 2)) %
         998244353;
}
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long int rng(long long int lim, long long int s = 0) {
  uniform_int_distribution<long long int> uid(s, lim - 1);
  return uid(rang);
}
double sqr(double n) {
  double s = 0, e = n;
  for (long long int i = 0; i < 50; i++) {
    double m = s + (e - s) / 2;
    if (m * m <= n)
      s = m;
    else
      e = m;
  }
  return s;
}
void pre_input() { fact_pre(); }
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  if (a[0] > a[1] + 1) {
    cout << 0 << "\n";
    return;
  }
  if (a[0] == a[1]) {
    cout << fact[n] << "\n";
    return;
  }
  long long int h = 0;
  for (auto &t : a)
    if (t == a[1]) h++;
  long long int k = n - h - 1;
  long long int ans = fact[n];
  for (long long int i = k; i >= 0; i--) {
    ans -= (((ncr(k, i) * fact[i]) % 998244353) * fact[n - i - 1]) % 998244353;
    ans = (ans + 998244353) % 998244353;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre_input();
  long long int tc;
  cin >> tc;
  for (long long int number = 1; number <= tc; number++) {
    solve();
  }
  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;
long long int power(long long int x, long long int b, long long int p = 1) {
  x %= 998244353;
  while (b > 0) {
    if (b & 1) p = (p * x) % 998244353;
    b >>= 1;
    x = (x * x) % 998244353;
  }
  return p;
}
vector<long long int> fact;
void fact_pre(long long int n = 1100000) {
  fact.resize(n, 1);
  for (long long int i = 2; i < n; i++) fact[i] = (fact[i - 1] * i) % 998244353;
}
long long int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  return (fact[n] * power((fact[n - r] * fact[r]) % 998244353, 998244353 - 2)) %
         998244353;
}
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long int rng(long long int lim, long long int s = 0) {
  uniform_int_distribution<long long int> uid(s, lim - 1);
  return uid(rang);
}
double sqr(double n) {
  double s = 0, e = n;
  for (long long int i = 0; i < 50; i++) {
    double m = s + (e - s) / 2;
    if (m * m <= n)
      s = m;
    else
      e = m;
  }
  return s;
}
void pre_input() { fact_pre(); }
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  if (a[0] > a[1] + 1) {
    cout << 0 << "\n";
    return;
  }
  if (a[0] == a[1]) {
    cout << fact[n] << "\n";
    return;
  }
  long long int h = 0;
  for (auto &t : a)
    if (t == a[1]) h++;
  long long int k = n - h - 1;
  long long int ans = fact[n];
  for (long long int i = k; i >= 0; i--) {
    ans -= (((ncr(k, i) * fact[i]) % 998244353) * fact[n - i - 1]) % 998244353;
    ans = (ans + 998244353) % 998244353;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre_input();
  long long int tc;
  cin >> tc;
  for (long long int number = 1; number <= tc; number++) {
    solve();
  }
  return 0;
}

```