#include <bits/stdc++.h>
using namespace std;
long long int modAdd(long long int a, long long int b) {
  return (a % 998244353 + b % 998244353) % 998244353;
}
long long int modSubtract(long long int a, long long int b) {
  return ((a % 998244353 - b % 998244353) + 998244353) % 998244353;
}
long long int modProduct(long long int a, long long int b) {
  return (a % 998244353 * b % 998244353) % 998244353;
}
long long int powerMod(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int x = powerMod(a, b / 2);
  if (b % 2 == 0)
    return (x * x) % 998244353;
  else
    return (((x * x) % 998244353) * a) % 998244353;
}
long long int modInverse(long long int a) {
  long long int x = powerMod(a, 998244353 - 2);
  return x;
}
long long int modDivide(long long int a, long long int b) {
  return (a % 998244353 * modInverse(b) % 998244353) % 998244353;
}
long long int fac[200005];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  for (long long int i = 1; i < 200005; i++) fac[i] = modProduct(fac[i - 1], i);
  long long int t;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    long long int n;
    cin >> n;
    long long int a[n], b[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = a[i];
    }
    sort(b, b + n);
    if (b[n - 1] - b[n - 2] > 1)
      cout << 0;
    else {
      if (b[n - 1] - b[n - 2] == 0)
        cout << fac[n];
      else {
        long long int count = 0;
        for (long long int i = n - 2; i >= 0; i--) {
          if (b[i] != b[n - 2]) break;
          count++;
        }
        long long int ans = 0;
        ans = modProduct(count, fac[n - 1]);
        long long int x = n - 2, y = count, z = n - count;
        for (long long int i = count + 1; i < n; i++) {
          ans = modAdd(ans, modProduct(n - i, modProduct(y, fac[x])));
          x--;
          z--;
          y = modProduct(y, z);
        }
        cout << ans;
      }
    }
    cout << endl;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
}

### 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;
long long int modAdd(long long int a, long long int b) {
  return (a % 998244353 + b % 998244353) % 998244353;
}
long long int modSubtract(long long int a, long long int b) {
  return ((a % 998244353 - b % 998244353) + 998244353) % 998244353;
}
long long int modProduct(long long int a, long long int b) {
  return (a % 998244353 * b % 998244353) % 998244353;
}
long long int powerMod(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int x = powerMod(a, b / 2);
  if (b % 2 == 0)
    return (x * x) % 998244353;
  else
    return (((x * x) % 998244353) * a) % 998244353;
}
long long int modInverse(long long int a) {
  long long int x = powerMod(a, 998244353 - 2);
  return x;
}
long long int modDivide(long long int a, long long int b) {
  return (a % 998244353 * modInverse(b) % 998244353) % 998244353;
}
long long int fac[200005];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  for (long long int i = 1; i < 200005; i++) fac[i] = modProduct(fac[i - 1], i);
  long long int t;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    long long int n;
    cin >> n;
    long long int a[n], b[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = a[i];
    }
    sort(b, b + n);
    if (b[n - 1] - b[n - 2] > 1)
      cout << 0;
    else {
      if (b[n - 1] - b[n - 2] == 0)
        cout << fac[n];
      else {
        long long int count = 0;
        for (long long int i = n - 2; i >= 0; i--) {
          if (b[i] != b[n - 2]) break;
          count++;
        }
        long long int ans = 0;
        ans = modProduct(count, fac[n - 1]);
        long long int x = n - 2, y = count, z = n - count;
        for (long long int i = count + 1; i < n; i++) {
          ans = modAdd(ans, modProduct(n - i, modProduct(y, fac[x])));
          x--;
          z--;
          y = modProduct(y, z);
        }
        cout << ans;
      }
    }
    cout << endl;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
}

```