#include <bits/stdc++.h>
using namespace std;
vector<long long int> fact(200005, 1);
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
          p) %
         p;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int m = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    m = max(m, a[i]);
  }
  long long int mx = 0, d = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] == m) {
      mx++;
    }
    if (abs(a[i] - m) > 1) {
      d++;
    }
  }
  if (mx > 1) {
    cout << fact[n];
    return;
  }
  if (d + mx == n) {
    cout << 0;
    return;
  }
  long long int ans = 0;
  for (long long int i = 0; i <= d; i++) {
    ans +=
        ((fact[n - 1 - i] % 998244353) *
         (fact[i] % 998244353 * nCrModPFermat(d, i, 998244353) % 998244353)) %
        998244353;
  }
  long long int x = (fact[n] % 998244353 - ans % 998244353) % 998244353;
  if (x < 0) x += 998244353;
  cout << x;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= 200005; i++) {
    fact[i] = (fact[i - 1] % 998244353 * i) % 998244353;
  }
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<long long int> fact(200005, 1);
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
          p) %
         p;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int m = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    m = max(m, a[i]);
  }
  long long int mx = 0, d = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] == m) {
      mx++;
    }
    if (abs(a[i] - m) > 1) {
      d++;
    }
  }
  if (mx > 1) {
    cout << fact[n];
    return;
  }
  if (d + mx == n) {
    cout << 0;
    return;
  }
  long long int ans = 0;
  for (long long int i = 0; i <= d; i++) {
    ans +=
        ((fact[n - 1 - i] % 998244353) *
         (fact[i] % 998244353 * nCrModPFermat(d, i, 998244353) % 998244353)) %
        998244353;
  }
  long long int x = (fact[n] % 998244353 - ans % 998244353) % 998244353;
  if (x < 0) x += 998244353;
  cout << x;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= 200005; i++) {
    fact[i] = (fact[i - 1] % 998244353 * i) % 998244353;
  }
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

```