#include <bits/stdc++.h>
using namespace std;
const int INF = 1000 * 1000 * 1000;
const long long LINF = 1LL * INF * INF;
const int MAX = 100010;
const long double PI = acos(-1.);
const double EPS = 1e-6;
const long long MOD = 998244353;
long long add(long long a, long long b) {
  long long f = a + b;
  if (f > MOD) f -= MOD;
  return f;
}
long long sub(long long a, long long b) {
  long long f = a - b;
  if (f < 0) f += MOD;
  return f;
}
long long mult(long long a, long long b) { return (a * b) % MOD; }
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) return mult(a, binpow(a, n - 1));
  long long b = binpow(a, n / 2);
  return mult(b, b);
}
long long inv(long long x) { return binpow(x, MOD - 2); }
long long fast_inv(long long x, long long m) {
  long long u = 0, v = 1;
  while (x != 0) {
    long long t = m / x;
    m -= t * x;
    swap(x, m);
    u -= t * v;
    swap(u, v);
  }
  return u;
}
long long divide(long long a, long long b) { return mult(a, inv(b)); }
long long dv(long long a, long long b) { return mult(a, fast_inv(b, MOD)); }
long long fact[200500];
long long C(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  }
  return divide(fact[n], mult(fact[k], fact[n - k]));
}
void prec() {
  fact[0] = 1;
  for (int i = 1; i < 200500; ++i) {
    fact[i] = mult(fact[i - 1], i);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto& a : v) cin >> a;
  ;
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    m[v[i]]++;
  }
  if (m.size() == 1) {
    cout << fact[n] << endl;
    return;
  }
  if (m.rbegin()->first - 1 > next(m.rbegin())->first &&
      m.rbegin()->second == 1) {
    cout << 0 << endl;
    return;
  }
  if (m.rbegin()->second > 1) {
    cout << fact[n] << endl;
    return;
  } else {
    int k = next(m.rbegin())->second;
    cout << sub(fact[n], mult(fact[k], mult(fact[n - k - 1], C(n, k + 1))))
         << endl;
  }
}
int main() {
  prec();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 INF = 1000 * 1000 * 1000;
const long long LINF = 1LL * INF * INF;
const int MAX = 100010;
const long double PI = acos(-1.);
const double EPS = 1e-6;
const long long MOD = 998244353;
long long add(long long a, long long b) {
  long long f = a + b;
  if (f > MOD) f -= MOD;
  return f;
}
long long sub(long long a, long long b) {
  long long f = a - b;
  if (f < 0) f += MOD;
  return f;
}
long long mult(long long a, long long b) { return (a * b) % MOD; }
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) return mult(a, binpow(a, n - 1));
  long long b = binpow(a, n / 2);
  return mult(b, b);
}
long long inv(long long x) { return binpow(x, MOD - 2); }
long long fast_inv(long long x, long long m) {
  long long u = 0, v = 1;
  while (x != 0) {
    long long t = m / x;
    m -= t * x;
    swap(x, m);
    u -= t * v;
    swap(u, v);
  }
  return u;
}
long long divide(long long a, long long b) { return mult(a, inv(b)); }
long long dv(long long a, long long b) { return mult(a, fast_inv(b, MOD)); }
long long fact[200500];
long long C(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  }
  return divide(fact[n], mult(fact[k], fact[n - k]));
}
void prec() {
  fact[0] = 1;
  for (int i = 1; i < 200500; ++i) {
    fact[i] = mult(fact[i - 1], i);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto& a : v) cin >> a;
  ;
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    m[v[i]]++;
  }
  if (m.size() == 1) {
    cout << fact[n] << endl;
    return;
  }
  if (m.rbegin()->first - 1 > next(m.rbegin())->first &&
      m.rbegin()->second == 1) {
    cout << 0 << endl;
    return;
  }
  if (m.rbegin()->second > 1) {
    cout << fact[n] << endl;
    return;
  } else {
    int k = next(m.rbegin())->second;
    cout << sub(fact[n], mult(fact[k], mult(fact[n - k - 1], C(n, k + 1))))
         << endl;
  }
}
int main() {
  prec();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```