#include <bits/stdc++.h>
using namespace std;
long long orz = 998244353LL;
int a[200005];
long long fac[200005] = {1, 1}, inv[200005] = {1, 1}, f[200005] = {1, 1};
void genfact() {
  for (int i = 2; i < 200005; i++) {
    fac[i] = fac[i - 1] * i % orz;
    f[i] = (orz - orz / i) * f[orz % i] % orz;
    inv[i] = inv[i - 1] * f[i] % orz;
  }
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * inv[m] % orz * inv[n - m] % orz;
}
int main() {
  int T, i, n, j;
  long long ans, cnt, sub;
  scanf("%d", &T);
  genfact();
  while (T--) {
    scanf("%d", &n);
    ans = fac[n];
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    for (i = n - 1, cnt = 0; i >= 0; i--) {
      if (a[i] == a[n - 1]) cnt++;
    }
    if (cnt > 1) {
      printf("%lld\n", ans);
      continue;
    }
    for (i = 0, cnt = 0; i < n; i++) {
      if (a[i] == a[n - 1] - 1) cnt++;
    }
    sub = ((C(n, cnt + 1) * fac[cnt] % orz) * fac[n - cnt - 1]) % orz;
    ans = (ans + orz - sub) % orz;
    printf("%lld\n", ans);
  }
  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 orz = 998244353LL;
int a[200005];
long long fac[200005] = {1, 1}, inv[200005] = {1, 1}, f[200005] = {1, 1};
void genfact() {
  for (int i = 2; i < 200005; i++) {
    fac[i] = fac[i - 1] * i % orz;
    f[i] = (orz - orz / i) * f[orz % i] % orz;
    inv[i] = inv[i - 1] * f[i] % orz;
  }
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * inv[m] % orz * inv[n - m] % orz;
}
int main() {
  int T, i, n, j;
  long long ans, cnt, sub;
  scanf("%d", &T);
  genfact();
  while (T--) {
    scanf("%d", &n);
    ans = fac[n];
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    for (i = n - 1, cnt = 0; i >= 0; i--) {
      if (a[i] == a[n - 1]) cnt++;
    }
    if (cnt > 1) {
      printf("%lld\n", ans);
      continue;
    }
    for (i = 0, cnt = 0; i < n; i++) {
      if (a[i] == a[n - 1] - 1) cnt++;
    }
    sub = ((C(n, cnt + 1) * fac[cnt] % orz) * fac[n - cnt - 1]) % orz;
    ans = (ans + orz - sub) % orz;
    printf("%lld\n", ans);
  }
  return 0;
}

```