#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ";
  err(++it, args...);
}
const long long MX = 2e5 + 105;
const long long MOD = 998244353;
inline void normal(long long &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1LL) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
long long fact[MX];
long long ara[MX];
void factorial() {
  fact[0] = 1;
  for (long long i = 1; i < MX; i++) fact[i] = modMul(fact[i - 1], i);
}
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  long long ans = modMul(fact[r], fact[n - r]);
  ans = modDiv(fact[n], ans);
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ara[i];
  }
  sort(ara, ara + n);
  if (ara[n - 2] == ara[n - 1]) {
    cout << fact[n] << endl;
    return;
  }
  if (ara[n - 1] - ara[n - 2] > 1) {
    cout << 0 << endl;
    return;
  }
  long long c = 1;
  for (int i = n - 3; i >= 0; i--) {
    if (ara[i] == ara[i + 1])
      c++;
    else
      break;
  }
  long long ans = ncr(n, c + 1);
  ans = modMul(ans, fact[n - c - 1]);
  ans = modMul(ans, fact[c]);
  ans = modSub(fact[n], ans);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  factorial();
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ";
  err(++it, args...);
}
const long long MX = 2e5 + 105;
const long long MOD = 998244353;
inline void normal(long long &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1LL) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
long long fact[MX];
long long ara[MX];
void factorial() {
  fact[0] = 1;
  for (long long i = 1; i < MX; i++) fact[i] = modMul(fact[i - 1], i);
}
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  long long ans = modMul(fact[r], fact[n - r]);
  ans = modDiv(fact[n], ans);
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ara[i];
  }
  sort(ara, ara + n);
  if (ara[n - 2] == ara[n - 1]) {
    cout << fact[n] << endl;
    return;
  }
  if (ara[n - 1] - ara[n - 2] > 1) {
    cout << 0 << endl;
    return;
  }
  long long c = 1;
  for (int i = n - 3; i >= 0; i--) {
    if (ara[i] == ara[i + 1])
      c++;
    else
      break;
  }
  long long ans = ncr(n, c + 1);
  ans = modMul(ans, fact[n - c - 1]);
  ans = modMul(ans, fact[c]);
  ans = modSub(fact[n], ans);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  factorial();
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```