#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
const ll mod = 998244353, inf = 1e18;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
}
vl fact(2e5 + 5, 1);
ll binPow(ll a, ll b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  ll ret = binPow(a, b / 2);
  if (b % 2 == 0) return (ret * ret) % mod;
  return ((ret * ret) % mod * a) % mod;
}
ll inv(ll a) { return (binPow(a, mod - 2) % mod + mod) % mod; }
ll binom(ll a, ll b) {
  if (b < 0 or a < 0) return 0;
  return (((fact[a] * inv(fact[b])) % mod * inv(fact[a - b])) % mod + mod) %
         mod;
}
int main() {
  setIO();
  int t;
  cin >> t;
  for (int i = 1; i <= 2e5; i++) fact[i] = (i * fact[i - 1]) % mod;
  while (t--) {
    int n;
    cin >> n;
    vi a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    sort(a.begin() + 1, a.end());
    if (a[n] == a[n - 1]) {
      cout << fact[n] << '\n';
      continue;
    }
    if (a[n] > a[n - 1] + 1) {
      cout << 0 << '\n';
      continue;
    }
    ll cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] == a[n - 1]) cnt++;
    }
    ll ans = fact[n];
    for (int i = 1; i <= n; i++) {
      if (i - 1 < cnt) continue;
      ll decrease =
          ((binom((i - 1), cnt) * fact[cnt]) % mod * fact[n - cnt - 1]) % mod;
      ans = (ans - decrease) % mod;
      if (ans < 0) ans += mod;
    }
    cout << ans << '\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;
using ll = long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
const ll mod = 998244353, inf = 1e18;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
}
vl fact(2e5 + 5, 1);
ll binPow(ll a, ll b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  ll ret = binPow(a, b / 2);
  if (b % 2 == 0) return (ret * ret) % mod;
  return ((ret * ret) % mod * a) % mod;
}
ll inv(ll a) { return (binPow(a, mod - 2) % mod + mod) % mod; }
ll binom(ll a, ll b) {
  if (b < 0 or a < 0) return 0;
  return (((fact[a] * inv(fact[b])) % mod * inv(fact[a - b])) % mod + mod) %
         mod;
}
int main() {
  setIO();
  int t;
  cin >> t;
  for (int i = 1; i <= 2e5; i++) fact[i] = (i * fact[i - 1]) % mod;
  while (t--) {
    int n;
    cin >> n;
    vi a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    sort(a.begin() + 1, a.end());
    if (a[n] == a[n - 1]) {
      cout << fact[n] << '\n';
      continue;
    }
    if (a[n] > a[n - 1] + 1) {
      cout << 0 << '\n';
      continue;
    }
    ll cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] == a[n - 1]) cnt++;
    }
    ll ans = fact[n];
    for (int i = 1; i <= n; i++) {
      if (i - 1 < cnt) continue;
      ll decrease =
          ((binom((i - 1), cnt) * fact[cnt]) % mod * fact[n - cnt - 1]) % mod;
      ans = (ans - decrease) % mod;
      if (ans < 0) ans += mod;
    }
    cout << ans << '\n';
  }
  return 0;
}

```