#include <bits/stdc++.h>
using namespace std;
long long modInverse(long long a, long long b) {
  return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
const long long mod = 998244353;
long long fact[222222];
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long K = 0; K < n; K++) {
    cin >> a[K];
  };
  long long mx = 0;
  sort(a, a + n);
  mx = a[n - 1];
  if (a[n - 1] - a[n - 2] > 1) {
    cout << 0 << "\n";
    return;
  }
  long long cnt_small = 0;
  for (int k = 0; k < n; k++) {
    if (a[k] < mx - 1) cnt_small++;
  }
  long long ans = fact[n];
  if (a[n - 2] == a[n - 1]) {
    cout << ans << "\n";
    return;
  }
  long long bad = 0;
  for (long long k = 0; k <= cnt_small; k++) {
    long long right =
        (fact[cnt_small] * modInverse(fact[cnt_small - k], mod)) % mod;
    bad = (bad + (right * fact[n - k - 1]) % mod) % mod;
  }
  ans = (ans - bad + mod) % mod;
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  fact[0] = 1;
  for (long long k = 1; k < 222222; k++) {
    fact[k] = (k * fact[k - 1]) % mod;
  }
  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;
long long modInverse(long long a, long long b) {
  return 1 < a ? b - modInverse(b % a, a) * b / a : 1;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
const long long mod = 998244353;
long long fact[222222];
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long K = 0; K < n; K++) {
    cin >> a[K];
  };
  long long mx = 0;
  sort(a, a + n);
  mx = a[n - 1];
  if (a[n - 1] - a[n - 2] > 1) {
    cout << 0 << "\n";
    return;
  }
  long long cnt_small = 0;
  for (int k = 0; k < n; k++) {
    if (a[k] < mx - 1) cnt_small++;
  }
  long long ans = fact[n];
  if (a[n - 2] == a[n - 1]) {
    cout << ans << "\n";
    return;
  }
  long long bad = 0;
  for (long long k = 0; k <= cnt_small; k++) {
    long long right =
        (fact[cnt_small] * modInverse(fact[cnt_small - k], mod)) % mod;
    bad = (bad + (right * fact[n - k - 1]) % mod) % mod;
  }
  ans = (ans - bad + mod) % mod;
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  fact[0] = 1;
  for (long long k = 1; k < 222222; k++) {
    fact[k] = (k * fact[k - 1]) % mod;
  }
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```