#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
long long F[N];
const long long M = 998244353;
long long amodb(long long a) { return a - (a / M * M); }
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = bigmod(a, b / 2);
  ans = amodb(ans * ans);
  if (b % 2) {
    ans = amodb(ans * a);
  }
  return ans;
}
void precal() {
  F[0] = 1;
  for (long long i = 1; i < N; i++) {
    F[i] = ((F[i - 1] % M) * (i % M)) % M;
  }
}
int main() {
  long long t = 1;
  precal();
  scanf("%lld", &t);
  while (t--) {
    long long n, i, j, k = 0, mx = 0, smx, cnt = 0, l, r, ans = 0;
    scanf("%lld", &n);
    vector<long long> v(n);
    for (i = 0; i < n; i++) {
      scanf("%lld", &v[i]);
    }
    sort(v.rbegin(), v.rend());
    if (v[0] - v[1] > 1) {
      printf("0\n");
    } else if (v[0] - v[1] == 0) {
      printf("%lld\n", F[n]);
    } else {
      for (i = 1; i < n; i++) {
        if (v[i] == v[1]) {
          k++;
        }
      }
      long long baad = (F[n] * bigmod(k + 1, M - 2)) % M;
      ans = (F[n] - baad) % M;
      ans += M;
      ans %= M;
      printf("%lld\n", ans);
    }
  }
}

### Prompt

Generate 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>
using namespace std;
const long long N = 2e5 + 5;
long long F[N];
const long long M = 998244353;
long long amodb(long long a) { return a - (a / M * M); }
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = bigmod(a, b / 2);
  ans = amodb(ans * ans);
  if (b % 2) {
    ans = amodb(ans * a);
  }
  return ans;
}
void precal() {
  F[0] = 1;
  for (long long i = 1; i < N; i++) {
    F[i] = ((F[i - 1] % M) * (i % M)) % M;
  }
}
int main() {
  long long t = 1;
  precal();
  scanf("%lld", &t);
  while (t--) {
    long long n, i, j, k = 0, mx = 0, smx, cnt = 0, l, r, ans = 0;
    scanf("%lld", &n);
    vector<long long> v(n);
    for (i = 0; i < n; i++) {
      scanf("%lld", &v[i]);
    }
    sort(v.rbegin(), v.rend());
    if (v[0] - v[1] > 1) {
      printf("0\n");
    } else if (v[0] - v[1] == 0) {
      printf("%lld\n", F[n]);
    } else {
      for (i = 1; i < n; i++) {
        if (v[i] == v[1]) {
          k++;
        }
      }
      long long baad = (F[n] * bigmod(k + 1, M - 2)) % M;
      ans = (F[n] - baad) % M;
      ans += M;
      ans %= M;
      printf("%lld\n", ans);
    }
  }
}

```