#include <bits/stdc++.h>
#pragma pack(1)
#pragma GCC target("avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1) % 998244353;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y);
long long int modInverse(long long int b, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long int modDivide(long long int a, long long int b, long long int m) {
  a = a % m;
  long long int inv = modInverse(b, m);
  return (inv * a) % m;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; ++i) cin >> a[i];
    sort(a, a + n);
    long long int cmax = 0;
    long long int csecmax = 0;
    long long int x = a[n - 1];
    for (long long int i = n - 1; i >= 0; --i) {
      if (a[i] != a[n - 1]) {
        x = a[i];
        break;
      }
    }
    for (long long int i = 0; i < n; ++i) {
      if (a[i] == a[n - 1])
        ++cmax;
      else if (a[i] == x)
        ++csecmax;
    }
    if (cmax > 1) {
      cout << fact(n) << endl;
    } else if ((a[n - 1] - x) > 1) {
      cout << "0" << endl;
    } else {
      long long int ans = 1;
      for (long long int i = csecmax + 1; i <= csecmax + cmax; ++i) {
        ans = ans * i % 998244353;
      }
      long long int fin =
          ((ans % 998244353) - (1 % 998244353) + 998244353) % 998244353;
      long long int z = ((fact(n) % 998244353) * (fin % 998244353)) % 998244353;
      cout << modDivide(z, ans, 998244353) << endl;
    }
  }
  return 0;
}

### 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>
#pragma pack(1)
#pragma GCC target("avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1) % 998244353;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y);
long long int modInverse(long long int b, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long int modDivide(long long int a, long long int b, long long int m) {
  a = a % m;
  long long int inv = modInverse(b, m);
  return (inv * a) % m;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; ++i) cin >> a[i];
    sort(a, a + n);
    long long int cmax = 0;
    long long int csecmax = 0;
    long long int x = a[n - 1];
    for (long long int i = n - 1; i >= 0; --i) {
      if (a[i] != a[n - 1]) {
        x = a[i];
        break;
      }
    }
    for (long long int i = 0; i < n; ++i) {
      if (a[i] == a[n - 1])
        ++cmax;
      else if (a[i] == x)
        ++csecmax;
    }
    if (cmax > 1) {
      cout << fact(n) << endl;
    } else if ((a[n - 1] - x) > 1) {
      cout << "0" << endl;
    } else {
      long long int ans = 1;
      for (long long int i = csecmax + 1; i <= csecmax + cmax; ++i) {
        ans = ans * i % 998244353;
      }
      long long int fin =
          ((ans % 998244353) - (1 % 998244353) + 998244353) % 998244353;
      long long int z = ((fact(n) % 998244353) * (fin % 998244353)) % 998244353;
      cout << modDivide(z, ans, 998244353) << endl;
    }
  }
  return 0;
}

```