#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000,1000000000")
using namespace std;
template <class T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (int i = 0; i < (int)vec.size(); ++i) is >> vec[i];
  return is;
}
template <class T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < (int)vec.size(); ++i) os << vec[i] << " ";
  return os;
}
template <class T>
istream& operator>>(istream& is, pair<T, T>& a) {
  is >> a.first >> a.second;
  return is;
}
const double pi = acos(-1.0);
const long long mod = 998244353;
const long long INF = 2e18;
const int inf = (int)1e9 + 7;
const int maxlen = (int)1e5 + 100;
const double eps = 1e-7;
long long binpow(long long a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
long long del(long long a, long long b) {
  long long d = binpow(b, mod - 2);
  return (a * d) % mod;
}
void solve() {
  int n;
  cin >> n;
  int mx = -1, mx_n, mn_n = -1;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a > mx) {
      if (a == mx + 1) {
        mn_n = mx_n;
      } else
        mn_n = 0;
      mx = a;
      mx_n = 0;
    }
    if (a == mx) mx_n++;
    if (a == mx - 1) mn_n++;
  }
  if (mx_n == 1 && mn_n == 0) {
    cout << 0 << endl;
    return;
  }
  if (mx_n > 1) {
    long long fact = 1;
    for (int i = 2; i <= n; i++) {
      fact = (fact * i) % mod;
    }
    cout << fact << endl;
    return;
  }
  long long fact = 1;
  vector<long long> f(n + 1);
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact = (fact * i) % mod;
    f[i] = fact;
  }
  long long sum = 0;
  for (int i = mn_n; i <= n - 1; i++) {
    long long jopa = del(f[i], f[i - mn_n]);
    jopa = (jopa * f[n - mn_n - 1]) % mod;
    sum = (sum + jopa) % mod;
  }
  sum = (f[n] + mod - sum) % mod;
  cout << sum << endl;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/STACK:1000000000,1000000000")
using namespace std;
template <class T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (int i = 0; i < (int)vec.size(); ++i) is >> vec[i];
  return is;
}
template <class T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < (int)vec.size(); ++i) os << vec[i] << " ";
  return os;
}
template <class T>
istream& operator>>(istream& is, pair<T, T>& a) {
  is >> a.first >> a.second;
  return is;
}
const double pi = acos(-1.0);
const long long mod = 998244353;
const long long INF = 2e18;
const int inf = (int)1e9 + 7;
const int maxlen = (int)1e5 + 100;
const double eps = 1e-7;
long long binpow(long long a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
long long del(long long a, long long b) {
  long long d = binpow(b, mod - 2);
  return (a * d) % mod;
}
void solve() {
  int n;
  cin >> n;
  int mx = -1, mx_n, mn_n = -1;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a > mx) {
      if (a == mx + 1) {
        mn_n = mx_n;
      } else
        mn_n = 0;
      mx = a;
      mx_n = 0;
    }
    if (a == mx) mx_n++;
    if (a == mx - 1) mn_n++;
  }
  if (mx_n == 1 && mn_n == 0) {
    cout << 0 << endl;
    return;
  }
  if (mx_n > 1) {
    long long fact = 1;
    for (int i = 2; i <= n; i++) {
      fact = (fact * i) % mod;
    }
    cout << fact << endl;
    return;
  }
  long long fact = 1;
  vector<long long> f(n + 1);
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact = (fact * i) % mod;
    f[i] = fact;
  }
  long long sum = 0;
  for (int i = mn_n; i <= n - 1; i++) {
    long long jopa = del(f[i], f[i - mn_n]);
    jopa = (jopa * f[n - mn_n - 1]) % mod;
    sum = (sum + jopa) % mod;
  }
  sum = (f[n] + mod - sum) % mod;
  cout << sum << endl;
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```