#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 9;
const long long inf = 0x3f3f3f3f;
const long long mod = 998244353;
long long a[N];
long long fac[N], inv_fac[N];
long long q_pow(long long a, long long b, long long mod) {
  long long ret = 1;
  for (long long x = a; b; b >>= 1, (x *= x) %= mod)
    if (b & 1) (ret *= x) %= mod;
  return ret;
}
long long inv(long long x) { return q_pow(x, mod - 2, mod); }
void init() {
  fac[0] = inv_fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = (fac[i - 1] * 1ll * i) % mod;
    inv_fac[i] = inv(fac[i]);
  }
}
long long C(long long n, long long m) {
  return fac[n] * inv_fac[m] % mod * inv_fac[n - m] % mod;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n, greater<int>());
  long long ma1 = 1, ma2 = 0;
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[1])
      ma1++;
    else if (a[i] + 1 == a[1])
      ma2++;
    else {
      if (ma2 == 0 && ma1 == 1) {
        cout << 0 << endl;
        return;
      }
      break;
    }
  }
  if (ma1 > 1) {
    cout << fac[n] << endl;
  } else {
    cout << fac[n] * inv_fac[ma2 + 1] % mod * ma2 % mod * fac[ma2] % mod
         << endl;
  }
}
signed main() {
  init();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create 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 = 1e6 + 9;
const long long inf = 0x3f3f3f3f;
const long long mod = 998244353;
long long a[N];
long long fac[N], inv_fac[N];
long long q_pow(long long a, long long b, long long mod) {
  long long ret = 1;
  for (long long x = a; b; b >>= 1, (x *= x) %= mod)
    if (b & 1) (ret *= x) %= mod;
  return ret;
}
long long inv(long long x) { return q_pow(x, mod - 2, mod); }
void init() {
  fac[0] = inv_fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = (fac[i - 1] * 1ll * i) % mod;
    inv_fac[i] = inv(fac[i]);
  }
}
long long C(long long n, long long m) {
  return fac[n] * inv_fac[m] % mod * inv_fac[n - m] % mod;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n, greater<int>());
  long long ma1 = 1, ma2 = 0;
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[1])
      ma1++;
    else if (a[i] + 1 == a[1])
      ma2++;
    else {
      if (ma2 == 0 && ma1 == 1) {
        cout << 0 << endl;
        return;
      }
      break;
    }
  }
  if (ma1 > 1) {
    cout << fac[n] << endl;
  } else {
    cout << fac[n] * inv_fac[ma2 + 1] % mod * ma2 % mod * fac[ma2] % mod
         << endl;
  }
}
signed main() {
  init();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```