#include <bits/stdc++.h>
using namespace std;
long long int m = 998244353;
struct hey {
  long long int x, y, g;
};
struct hey inv1(long long int a, long long int b) {
  if (b == 0) {
    struct hey ans;
    ans.g = a;
    ans.y = 0;
    ans.x = 1;
    return ans;
  } else {
    struct hey p = inv1(b, a % b);
    struct hey ans;
    ans.g = p.g;
    ans.x = p.y;
    ans.y = p.x - ((a / b) * (p.y));
    return ans;
  }
}
long long int inv(long long int b) {
  struct hey ans = inv1(b, m);
  return ((ans.x) + m) % m;
}
long long int bin(long long int a, long long int b, long long int fact[]) {
  if (b < 0 || a < 0) return 0;
  return (((fact[a] * inv(fact[b])) % m * inv(fact[a - b])) % m + m) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, n1 = 2e5, c, d, ans;
  long long int fact[n1 + 1];
  fact[0] = 1;
  for (long long int i = 1; i <= n1; i++) {
    fact[i] = (i * fact[i - 1]) % m;
  }
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    if (a[n - 1] > (a[n - 2] + 1)) {
      cout << "0\n";
      continue;
    }
    if (a[n - 1] == a[n - 2]) {
      cout << (fact[n]) << "\n";
      continue;
    }
    c = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == (a[n - 2])) c++;
    }
    ans = fact[n];
    for (int i = 0; i < n; i++) {
      if (i < c) continue;
      d = ((((bin(i, c, fact) * fact[c])) % m) * fact[n - c - 1]) % m;
      ans = (ans - d) % m;
      if (ans < 0) {
        ans = ans + m;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long int m = 998244353;
struct hey {
  long long int x, y, g;
};
struct hey inv1(long long int a, long long int b) {
  if (b == 0) {
    struct hey ans;
    ans.g = a;
    ans.y = 0;
    ans.x = 1;
    return ans;
  } else {
    struct hey p = inv1(b, a % b);
    struct hey ans;
    ans.g = p.g;
    ans.x = p.y;
    ans.y = p.x - ((a / b) * (p.y));
    return ans;
  }
}
long long int inv(long long int b) {
  struct hey ans = inv1(b, m);
  return ((ans.x) + m) % m;
}
long long int bin(long long int a, long long int b, long long int fact[]) {
  if (b < 0 || a < 0) return 0;
  return (((fact[a] * inv(fact[b])) % m * inv(fact[a - b])) % m + m) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, n1 = 2e5, c, d, ans;
  long long int fact[n1 + 1];
  fact[0] = 1;
  for (long long int i = 1; i <= n1; i++) {
    fact[i] = (i * fact[i - 1]) % m;
  }
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    if (a[n - 1] > (a[n - 2] + 1)) {
      cout << "0\n";
      continue;
    }
    if (a[n - 1] == a[n - 2]) {
      cout << (fact[n]) << "\n";
      continue;
    }
    c = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] == (a[n - 2])) c++;
    }
    ans = fact[n];
    for (int i = 0; i < n; i++) {
      if (i < c) continue;
      d = ((((bin(i, c, fact) * fact[c])) % m) * fact[n - c - 1]) % m;
      ans = (ans - d) % m;
      if (ans < 0) {
        ans = ans + m;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```