#include <bits/stdc++.h>
using namespace std;
template <typename Head>
inline void dout(Head in) {
  cerr << in << '\n';
}
long long bp(long long n, long long m, long long mod) {
  long long res = 1;
  while (m) {
    m & 1 ? res = res * n % mod : res;
    n = n * n % mod;
    m >>= 1LL;
  }
  return res;
}
int32_t main() {
  cin.tie(nullptr)->ios::sync_with_stdio(false);
  long long constexpr mod = (long long)998244353;
  long long constexpr maxn = (long long)2e5 + 1;
  vector<long long> fac(maxn);
  fac[0] = fac[1] = 1LL;
  for (long long i = 2; i < maxn; i++) {
    fac[i] = fac[i - 1] * (long long)i % mod;
  }
  vector<long long> inv(maxn);
  inv[maxn - 1] = bp(fac[maxn - 1], mod - 2, mod);
  for (long long i = maxn - 2; ~i; i--) {
    inv[i] = inv[i + 1] * (long long)(i + 1) % mod;
  }
  function<long long(long long, long long)> A = [&](long long l, long long r) {
    return fac[r] * inv[l] % mod;
  };
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> cnt;
    for (long long &e : a) {
      cin >> e;
      cnt[e]++;
    }
    sort(a.begin(), a.end());
    if (cnt[a.back()] > 1) {
      cout << fac[n] << "\n";
      continue;
    }
    if (a.back() - a[(long long)a.size() - 2] > 1) {
      cout << "0\n";
      continue;
    }
    long long ans = fac[n];
    ans = (ans - fac[cnt[a[(long long)a.size() - 2]]] *
                     A(cnt[a[(long long)a.size() - 2]] + 1, n)) %
          mod;
    cout << (ans + mod) % mod << "\n";
  }
  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;
template <typename Head>
inline void dout(Head in) {
  cerr << in << '\n';
}
long long bp(long long n, long long m, long long mod) {
  long long res = 1;
  while (m) {
    m & 1 ? res = res * n % mod : res;
    n = n * n % mod;
    m >>= 1LL;
  }
  return res;
}
int32_t main() {
  cin.tie(nullptr)->ios::sync_with_stdio(false);
  long long constexpr mod = (long long)998244353;
  long long constexpr maxn = (long long)2e5 + 1;
  vector<long long> fac(maxn);
  fac[0] = fac[1] = 1LL;
  for (long long i = 2; i < maxn; i++) {
    fac[i] = fac[i - 1] * (long long)i % mod;
  }
  vector<long long> inv(maxn);
  inv[maxn - 1] = bp(fac[maxn - 1], mod - 2, mod);
  for (long long i = maxn - 2; ~i; i--) {
    inv[i] = inv[i + 1] * (long long)(i + 1) % mod;
  }
  function<long long(long long, long long)> A = [&](long long l, long long r) {
    return fac[r] * inv[l] % mod;
  };
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> cnt;
    for (long long &e : a) {
      cin >> e;
      cnt[e]++;
    }
    sort(a.begin(), a.end());
    if (cnt[a.back()] > 1) {
      cout << fac[n] << "\n";
      continue;
    }
    if (a.back() - a[(long long)a.size() - 2] > 1) {
      cout << "0\n";
      continue;
    }
    long long ans = fac[n];
    ans = (ans - fac[cnt[a[(long long)a.size() - 2]]] *
                     A(cnt[a[(long long)a.size() - 2]] + 1, n)) %
          mod;
    cout << (ans + mod) % mod << "\n";
  }
  return 0;
}

```