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;
const int MOD = 998244353;
template <typename T>
T add(T x) {
  return x;
}
template <typename T, typename... Ts>
T add(T x, Ts... y) {
  T res = x + add(y...);
  if (res >= MOD) res -= MOD;
  return res;
}
template <typename T, typename... Ts>
T sub(T x, Ts... y) {
  return add(x, MOD - add(y...));
}
template <typename T, typename... Ts>
void udd(T &x, Ts... y) {
  x = add(x, y...);
}
template <typename T>
T mul(T x) {
  return x;
}
template <typename T, typename... Ts>
T mul(T x, Ts... y) {
  return (x * 1ll * mul(y...)) % MOD;
}
template <typename T, typename... Ts>
void uul(T &x, Ts... y) {
  x = mul(x, y...);
}
int bin(int a, long long deg) {
  int r = 1;
  while (deg) {
    if (deg & 1) uul(r, a);
    deg >>= 1;
    uul(a, a);
  }
  return r;
}
int inv(int x) {
  assert(x);
  return bin(x, MOD - 2);
}
vector<int> fac(2 * 1e5 + 1);
void precalc() {
  fac[0] = 1;
  for (int i = 1; i <= 2 * 1e5; ++i) {
    fac[i] = mul(fac[i - 1], i);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &val : a) {
    cin >> val;
  }
  int maxx = a[0];
  int maxx_cnt = 1;
  int second_max = -1;
  int second_max_cnt = 0;
  for (int i = 1; i < n; ++i) {
    if (a[i] > maxx) {
      second_max = maxx;
      second_max_cnt = maxx_cnt;
      maxx = a[i];
      maxx_cnt = 1;
    } else if (a[i] == maxx) {
      ++maxx_cnt;
    } else if (second_max < a[i] && a[i] < maxx) {
      second_max = a[i];
      second_max_cnt = 1;
    } else if (a[i] == second_max) {
      ++second_max_cnt;
    }
  }
  if (second_max == -1) {
    cout << fac[n] << "\n";
    return;
  }
  if (maxx_cnt == 2) {
    cout << fac[n] << "\n";
    return;
  }
  if (second_max != maxx - 1) {
    cout << "0\n";
    return;
  }
  cout << mul(fac[n], inv(second_max_cnt + 1), second_max_cnt) << "\n";
}
int main() {
  precalc();
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    solve();
  }
}
