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.
#include <bits/stdc++.h>
using namespace std;
long long md = 998244353;
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2);
  x = (x * x) % md;
  if (b % 2 == 1) x = (x * a) % md;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc;
  cin >> tc;
  long long fq[200005], i;
  fq[0] = 1;
  fq[1] = 1;
  for (i = 2; i <= 200005; i++) {
    fq[i] = fq[i - 1] * i;
    fq[i] %= md;
  }
  while (tc--) {
    long long n, an = 0, oo, a[200005], j, k, p, ans, ok = 0;
    cin >> n;
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    p = a[0];
    for (i = 0; i < n - 1; i++) {
      if (a[i] == p || a[i] == p + 1) {
        p = a[i];
      } else {
        ok = 1;
        break;
      }
    }
    if (ok) {
      p = a[n - 1];
      if (a[n - 2] == p) {
        cout << fq[n] << endl;
      } else
        cout << 0 << endl;
    } else {
      if (a[n - 1] == a[n - 2]) {
        cout << fq[n] << endl;
      } else {
        p = a[n - 1];
        p--;
        for (i = 0; i < n; i++) {
          if (a[i] < p) an++;
        }
        ans = fq[n] + md;
        ans -= (fq[n - 1]);
        for (i = 1; i <= an; i++) {
          ans += md;
          oo = (fq[n - i - 1]);
          oo %= md;
          oo = oo * fq[an];
          oo %= md;
          if (an != i) oo = oo * bigmod(an - i, md - 2);
          oo %= md;
          ans -= oo;
          ans += 3 * md;
          ans %= md;
        }
        ans %= md;
        cout << ans << endl;
      }
    }
  }
}
