#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int N = 1e6 + 4;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
vector<long long> first(N, 1), INV(N, 1), minp(N, 0);
vector<bool> prime(N, 0);
long long gd(long long a, long long b) {
  if (b == 0) return a;
  return gd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gd(a, b); }
long long ex(long long x, long long n) {
  long long res = 1;
  x = x % mod;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n = n >> 1;
  }
  return res % mod;
}
long long fermat(long long n) { return ex(n, mod - 2); }
void fact() {
  for (int i = 2; i < N; i++) {
    first[i] = (first[i - 1] * i) % mod;
    INV[i] = fermat(first[i]);
  }
}
long long ncr(long long n, long long r) {
  if (r == 0 || n == r) return 1;
  if (r > n) return 0;
  return (first[n] * INV[r] % mod * INV[n - r] % mod) % mod;
}
void sieve() {
  prime[0] = 1, prime[1] = 1;
  for (int i = 2; i * i < N; i++) {
    if (prime[i] == 0) {
      for (int j = i * i; j < N; j += i) prime[j] = 1;
    }
  }
}
void minprime() {
  minp[0] = 0, minp[1] = 1;
  for (int i = 2; i < N; i++) {
    if (minp[i] == 0) {
      minp[i] = i;
      for (int j = i * i; j < N; j += i) {
        if (minp[j] == 0) minp[j] = i;
      }
    }
  }
}
void JAI_SHREE_RAM() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  map<long long, long long> m;
  for (int i = 0; i < n; i++) {
    m[a[i]]++;
  }
  if (abs(a[n - 1] - a[n - 2]) >= 2) {
    cout << 0 << "\n";
    return;
  }
  if (abs(a[n - 1] - a[n - 2]) == 0) {
    cout << first[n] << "\n";
    return;
  }
  long long invalid = first[n] * fermat(m[a[n - 2]] + 1) % mod;
  long long ans = (first[n] - invalid + mod) % mod;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fact();
  long long tttt = 1;
  cin >> tttt;
  while (tttt--) {
    JAI_SHREE_RAM();
  }
}

### Prompt

Generate a Cpp solution 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;
const int mod = 998244353;
const int N = 1e6 + 4;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
vector<long long> first(N, 1), INV(N, 1), minp(N, 0);
vector<bool> prime(N, 0);
long long gd(long long a, long long b) {
  if (b == 0) return a;
  return gd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gd(a, b); }
long long ex(long long x, long long n) {
  long long res = 1;
  x = x % mod;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n = n >> 1;
  }
  return res % mod;
}
long long fermat(long long n) { return ex(n, mod - 2); }
void fact() {
  for (int i = 2; i < N; i++) {
    first[i] = (first[i - 1] * i) % mod;
    INV[i] = fermat(first[i]);
  }
}
long long ncr(long long n, long long r) {
  if (r == 0 || n == r) return 1;
  if (r > n) return 0;
  return (first[n] * INV[r] % mod * INV[n - r] % mod) % mod;
}
void sieve() {
  prime[0] = 1, prime[1] = 1;
  for (int i = 2; i * i < N; i++) {
    if (prime[i] == 0) {
      for (int j = i * i; j < N; j += i) prime[j] = 1;
    }
  }
}
void minprime() {
  minp[0] = 0, minp[1] = 1;
  for (int i = 2; i < N; i++) {
    if (minp[i] == 0) {
      minp[i] = i;
      for (int j = i * i; j < N; j += i) {
        if (minp[j] == 0) minp[j] = i;
      }
    }
  }
}
void JAI_SHREE_RAM() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  map<long long, long long> m;
  for (int i = 0; i < n; i++) {
    m[a[i]]++;
  }
  if (abs(a[n - 1] - a[n - 2]) >= 2) {
    cout << 0 << "\n";
    return;
  }
  if (abs(a[n - 1] - a[n - 2]) == 0) {
    cout << first[n] << "\n";
    return;
  }
  long long invalid = first[n] * fermat(m[a[n - 2]] + 1) % mod;
  long long ans = (first[n] - invalid + mod) % mod;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fact();
  long long tttt = 1;
  cin >> tttt;
  while (tttt--) {
    JAI_SHREE_RAM();
  }
}

```