#include <bits/stdc++.h>
using namespace std;
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
vector<long long> to_binary(long long n) {
  int i = 7;
  vector<long long> ans(8);
  while (n >= 2) {
    ans[i] = n % 2;
    n /= 2;
    i--;
  }
  ans[i] = n;
  return ans;
}
map<long long, long long> primefactorisation(long long n) {
  map<long long, long long> mp;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      mp[i] = 0;
      while (n % i == 0) {
        mp[i]++;
        n /= i;
      }
    }
  }
  if (n > 1) {
    mp[n] = 1;
  }
  return mp;
}
vector<bool> seive(long long n) {
  vector<bool> arr(n + 1, true);
  for (int i = 2; i * i <= n; i++) {
    if (arr[i]) {
      for (int j = i * i; j <= n; j += i) {
        arr[j] = false;
      }
    }
  }
  return arr;
}
long long powM(long long x, long long y) {
  long long ans = 1, r = 1;
  x %= (long long)(998244353);
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= (long long)(998244353);
    }
    r <<= 1;
    x *= x;
    x %= (long long)(998244353);
  }
  return ans;
}
bool isprime(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return 0;
    }
  }
  return 1;
}
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return (p1.first > p2.first);
}
void ipgraph(vector<set<long long>>& edge, long long v, long long e) {
  for (int i = 0; i < e; i++) {
    long long a, b;
    cin >> a >> b;
    edge[a].insert(b);
    edge[b].insert(a);
  }
}
void bfs(vector<vector<long long>>& edge, long long start,
         vector<bool>& visited) {
  long long curr = start;
  queue<long long> qq;
  qq.push(curr);
  visited[curr] = 1;
  while (!qq.empty()) {
    curr = qq.front();
    for (long long x : edge[curr]) {
      if (!visited[x]) {
        qq.push(x);
        visited[x] = 1;
      }
    }
    qq.pop();
  }
}
vector<string> bor = {"YES", "NO"};
vector<long long> fact(300001);
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < fact.size(); i++) {
    fact[i] = mod_mul(i, fact[i - 1], (long long)(998244353));
  }
}
unsigned long long power(unsigned long long x, int y, int p) {
  unsigned long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long modInverse(unsigned long long n, int p) {
  return power(n, p - 2, p);
}
unsigned long long ncr(unsigned long long n, int r, int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
          p) %
         p;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr.begin(), arr.end(), greater<long long>());
  long long maxx = arr[0];
  if (n == 1) {
    cout << 1 << "\n";
    return;
  }
  if (arr[1] == arr[0]) {
    cout << fact[n] % (long long)(998244353) << "\n";
    return;
  }
  if (arr[0] - arr[1] > 1) {
    cout << 0 << "\n";
    return;
  }
  long long x = 0;
  for (int i = 1; i < n; i++) {
    if (arr[i] == arr[1])
      x++;
    else
      break;
  }
  long long nf = fact[n];
  long long k = fact[n - x - 1];
  long long np = 0;
  long long diff = 1;
  long long ans = fact[n];
  for (int i = x + 1; i <= n; i++) {
    long long temp = ncr(i - 1, x, (long long)(998244353));
    temp = mod_mul(temp, fact[x], (long long)(998244353));
    temp = mod_mul(temp, k, (long long)(998244353));
    ans = mod_sub(ans, temp, (long long)(998244353));
    diff++;
  }
  cout << ans % (long long)(998244353) << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  pre();
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
vector<long long> to_binary(long long n) {
  int i = 7;
  vector<long long> ans(8);
  while (n >= 2) {
    ans[i] = n % 2;
    n /= 2;
    i--;
  }
  ans[i] = n;
  return ans;
}
map<long long, long long> primefactorisation(long long n) {
  map<long long, long long> mp;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      mp[i] = 0;
      while (n % i == 0) {
        mp[i]++;
        n /= i;
      }
    }
  }
  if (n > 1) {
    mp[n] = 1;
  }
  return mp;
}
vector<bool> seive(long long n) {
  vector<bool> arr(n + 1, true);
  for (int i = 2; i * i <= n; i++) {
    if (arr[i]) {
      for (int j = i * i; j <= n; j += i) {
        arr[j] = false;
      }
    }
  }
  return arr;
}
long long powM(long long x, long long y) {
  long long ans = 1, r = 1;
  x %= (long long)(998244353);
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= (long long)(998244353);
    }
    r <<= 1;
    x *= x;
    x %= (long long)(998244353);
  }
  return ans;
}
bool isprime(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return 0;
    }
  }
  return 1;
}
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return (p1.first > p2.first);
}
void ipgraph(vector<set<long long>>& edge, long long v, long long e) {
  for (int i = 0; i < e; i++) {
    long long a, b;
    cin >> a >> b;
    edge[a].insert(b);
    edge[b].insert(a);
  }
}
void bfs(vector<vector<long long>>& edge, long long start,
         vector<bool>& visited) {
  long long curr = start;
  queue<long long> qq;
  qq.push(curr);
  visited[curr] = 1;
  while (!qq.empty()) {
    curr = qq.front();
    for (long long x : edge[curr]) {
      if (!visited[x]) {
        qq.push(x);
        visited[x] = 1;
      }
    }
    qq.pop();
  }
}
vector<string> bor = {"YES", "NO"};
vector<long long> fact(300001);
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < fact.size(); i++) {
    fact[i] = mod_mul(i, fact[i - 1], (long long)(998244353));
  }
}
unsigned long long power(unsigned long long x, int y, int p) {
  unsigned long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long modInverse(unsigned long long n, int p) {
  return power(n, p - 2, p);
}
unsigned long long ncr(unsigned long long n, int r, int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
          p) %
         p;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr.begin(), arr.end(), greater<long long>());
  long long maxx = arr[0];
  if (n == 1) {
    cout << 1 << "\n";
    return;
  }
  if (arr[1] == arr[0]) {
    cout << fact[n] % (long long)(998244353) << "\n";
    return;
  }
  if (arr[0] - arr[1] > 1) {
    cout << 0 << "\n";
    return;
  }
  long long x = 0;
  for (int i = 1; i < n; i++) {
    if (arr[i] == arr[1])
      x++;
    else
      break;
  }
  long long nf = fact[n];
  long long k = fact[n - x - 1];
  long long np = 0;
  long long diff = 1;
  long long ans = fact[n];
  for (int i = x + 1; i <= n; i++) {
    long long temp = ncr(i - 1, x, (long long)(998244353));
    temp = mod_mul(temp, fact[x], (long long)(998244353));
    temp = mod_mul(temp, k, (long long)(998244353));
    ans = mod_sub(ans, temp, (long long)(998244353));
    diff++;
  }
  cout << ans % (long long)(998244353) << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  pre();
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```