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.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll mod = 998244353;
inline namespace {
template <typename B, typename E>
inline B slow_pow(const B &b, const E &e) {
  assert(e >= 0);
  B res = 1;
  for (int q = 0; q < e; q++) res *= b;
  return res;
}
template <typename B, typename E>
inline B bin_pow(const B &b, const E &e) {
  assert(e >= 0);
  B res = 1;
  B a = b;
  for (E q = e; q; q >>= 1) {
    if (q & 1) res *= a;
    a *= a;
  }
  return res;
}
template <typename B, typename E>
inline B mod_pow(const B &b, const E &e) {
  assert(e >= 0);
  B res = 1;
  B a = b;
  for (E q = e; q; q >>= 1) {
    if (q & 1) res = (res * a) % mod;
    a = (a * a) % mod;
  }
  return res;
}
}  // namespace
ll fact[200001];
void pre() {
  fact[0] = 1;
  for (ll x = 1; x <= 200000; x++) fact[x] = (fact[x - 1] * x) % mod;
}
ll mod_inv(const ll &x) { return mod_pow(x, mod - 2); }
ll choose(const int &n, const int &k) {
  return (((fact[n] * mod_inv(fact[k])) % mod) * mod_inv(fact[n - k])) % mod;
}
void solve() {
  int n;
  cin >> n;
  vector<ll> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  ll ans;
  if (a[n - 1] == a[n - 2])
    ans = fact[n];
  else if (a[n - 1] == a[n - 2] + 1) {
    int cnt = 0;
    ans = 0;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] + 1 == a[n - 1]) cnt++;
    }
    ans = (((choose(n, cnt + 1) * (fact[cnt + 1] - fact[cnt])) % mod) *
           fact[n - cnt - 1]) %
          mod;
  } else
    ans = 0;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  pre();
  int T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}
