description: You are given n integers a_1, a_2, …, a_n and an integer k. Find the maximum value of i ⋅ j - k ⋅ (a_i | a_j) over all pairs (i, j) of integers with 1 ≤ i < j ≤ n. Here, | is the [bitwise OR operator](https://en.wikipedia.org/wiki/Bitwise_operation#OR).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.
  
  The first line of each test case contains two integers n (2 ≤ n ≤ 10^5) and k (1 ≤ k ≤ min(n, 100)).
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print a single integer — the maximum possible value of i ⋅ j - k ⋅ (a_i | a_j).
  
  Example
  
  Input
  
  
  4
  3 3
  1 1 3
  2 2
  1 2
  4 3
  0 1 2 3
  6 6
  3 2 0 0 5 6
  
  
  Output
  
  
  -1
  -4
  3
  12
  
  Note
  
  Let f(i, j) = i ⋅ j - k ⋅ (a_i | a_j).
  
  In the first test case, 
  
    * f(1, 2) = 1 ⋅ 2 - k ⋅ (a_1 | a_2) = 2 - 3 ⋅ (1 | 1) = -1. 
    * f(1, 3) = 1 ⋅ 3 - k ⋅ (a_1 | a_3) = 3 - 3 ⋅ (1 | 3) = -6. 
    * f(2, 3) = 2 ⋅ 3 - k ⋅ (a_2 | a_3) = 6 - 3 ⋅ (1 | 3) = -3. 
  
  
  
  So the maximum is f(1, 2) = -1.
  
  In the fourth test case, the maximum is f(3, 4) = 12.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  void __p(T a) {
    cout << a;
  }
  template <typename T, typename U>
  void __p(pair<T, U> a) {
    cout << "{";
    __p(a.first);
    cout << ",";
    __p(a.second);
    cout << "}";
  }
  template <typename T>
  void __p(std::vector<T> a) {
    cout << "{";
    for (auto it = a.begin(); it < a.end(); it++)
      __p(*it), cout << ",}"[it + 1 == a.end()];
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    if (n == 2) {
      cout << 2 - k * (a[0] | a[1]) << endl;
      return;
    }
    long long ans = 2 - k * (a[0] | a[1]);
    for (long long i = n - 2; i > n - 300 && i >= 0; i--) {
      for (long long j = n - 1; j > i; j--) {
        ans = max(ans, (i + 1) * (j + 1) - k * (a[i] | a[j]));
      }
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
      solve();
    }
  }
  