#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll KMOD = 998244353;
ll mmul(ll a, ll b) { return a * b % KMOD; }
ll extended_euclidean_algorithm(ll x, ll y, ll n_ax, ll n_mx, ll n_ay,
                                ll n_my) {
  if (y == 1) return n_ay;
  ll rem = x % y;
  ll q = x / y;
  return extended_euclidean_algorithm(y, rem, n_ay, n_my, n_ax - n_ay * q,
                                      n_mx - n_my * q);
}
ll ModInverse(ll a, ll m) {
  return (extended_euclidean_algorithm(m, a % m, 0, 1, 1, 0) + m) % m;
}
ll mdiv(ll a, ll b) { return mmul(a, ModInverse(b, KMOD)); }
vector<ll> factorial;
void Preprocess() {
  factorial.resize(200001);
  factorial[0] = 1;
  factorial[1] = 1;
  for (int i = 2; i <= 200000; ++i) {
    factorial[i] = mmul(factorial[i - 1], i);
  }
}
int Solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int max = *max_element((a).begin(), (a).end()), maxcnt = 0, max_m1_cnt = 0;
  for (int x : a) {
    if (x == max) {
      maxcnt++;
    } else if (x == max - 1) {
      max_m1_cnt++;
    }
  }
  if (maxcnt > 1) {
    return factorial[n];
  } else if (max_m1_cnt == 0) {
    return 0;
  } else {
    return mmul(factorial[n], mdiv(max_m1_cnt, max_m1_cnt + 1));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  Preprocess();
  int t;
  cin >> t;
  while (t--) cout << Solve() << "\n";
}

### Prompt

Please create a solution in CPP 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;
using ll = long long;
ll KMOD = 998244353;
ll mmul(ll a, ll b) { return a * b % KMOD; }
ll extended_euclidean_algorithm(ll x, ll y, ll n_ax, ll n_mx, ll n_ay,
                                ll n_my) {
  if (y == 1) return n_ay;
  ll rem = x % y;
  ll q = x / y;
  return extended_euclidean_algorithm(y, rem, n_ay, n_my, n_ax - n_ay * q,
                                      n_mx - n_my * q);
}
ll ModInverse(ll a, ll m) {
  return (extended_euclidean_algorithm(m, a % m, 0, 1, 1, 0) + m) % m;
}
ll mdiv(ll a, ll b) { return mmul(a, ModInverse(b, KMOD)); }
vector<ll> factorial;
void Preprocess() {
  factorial.resize(200001);
  factorial[0] = 1;
  factorial[1] = 1;
  for (int i = 2; i <= 200000; ++i) {
    factorial[i] = mmul(factorial[i - 1], i);
  }
}
int Solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int max = *max_element((a).begin(), (a).end()), maxcnt = 0, max_m1_cnt = 0;
  for (int x : a) {
    if (x == max) {
      maxcnt++;
    } else if (x == max - 1) {
      max_m1_cnt++;
    }
  }
  if (maxcnt > 1) {
    return factorial[n];
  } else if (max_m1_cnt == 0) {
    return 0;
  } else {
    return mmul(factorial[n], mdiv(max_m1_cnt, max_m1_cnt + 1));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  Preprocess();
  int t;
  cin >> t;
  while (t--) cout << Solve() << "\n";
}

```