#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 998244353;
int fact[N], invfact[N];
int power(int x, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = (1LL * ret * x) % M;
    x = (1LL * x * x) % M;
    p >>= 1;
  }
  return ret;
}
int inv(int x) { return power(x, M - 2); }
void go() {
  fact[0] = 1;
  invfact[0] = inv(fact[0]);
  for (int i = 1; i < N; i++) {
    fact[i] = (1LL * fact[i - 1] * i) % M;
    invfact[i] = inv(fact[i]);
  }
}
int nCr(int n, int c) {
  if (c > n) return 0;
  return 1LL * fact[n] * invfact[n - c] % M * invfact[c] % M;
}
int main() {
  go();
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<int> arr(n);
    for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
    sort(arr.rbegin(), arr.rend());
    if (arr[0] == arr[1]) {
      printf("%d\n", fact[n]);
      continue;
    }
    if (arr[0] - arr[1] > 1) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
      if (arr[1] == arr[i]) ++cnt;
    }
    int res = 0;
    for (int i = 1; i < n; i++) {
      res = (res + fact[n - 1]) % M;
      res =
          (res -
           (1LL * nCr(n - cnt - 1, n - i) * fact[i - 1] % M * fact[n - i] % M) +
           M) %
          M;
    }
    printf("%d\n", res);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 2e5 + 5, M = 998244353;
int fact[N], invfact[N];
int power(int x, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = (1LL * ret * x) % M;
    x = (1LL * x * x) % M;
    p >>= 1;
  }
  return ret;
}
int inv(int x) { return power(x, M - 2); }
void go() {
  fact[0] = 1;
  invfact[0] = inv(fact[0]);
  for (int i = 1; i < N; i++) {
    fact[i] = (1LL * fact[i - 1] * i) % M;
    invfact[i] = inv(fact[i]);
  }
}
int nCr(int n, int c) {
  if (c > n) return 0;
  return 1LL * fact[n] * invfact[n - c] % M * invfact[c] % M;
}
int main() {
  go();
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<int> arr(n);
    for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
    sort(arr.rbegin(), arr.rend());
    if (arr[0] == arr[1]) {
      printf("%d\n", fact[n]);
      continue;
    }
    if (arr[0] - arr[1] > 1) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
      if (arr[1] == arr[i]) ++cnt;
    }
    int res = 0;
    for (int i = 1; i < n; i++) {
      res = (res + fact[n - 1]) % M;
      res =
          (res -
           (1LL * nCr(n - cnt - 1, n - i) * fact[i - 1] % M * fact[n - i] % M) +
           M) %
          M;
    }
    printf("%d\n", res);
  }
  return 0;
}

```