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;
  const long long oo = 1e18;
  const long long N = 1e6 + 50;
  const long long mod = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    int test_cases = 1;
    cin >> test_cases;
    while (test_cases--) {
      long long n, k;
      cin >> n >> k;
      long long a[n + 10];
      long long ans = -oo;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
      }
      for (long long i = 1; i < min(n - 1, 220LL); i++) {
        for (long long j = i + 1; j <= min(n, 220LL); j++) {
          ans = max(i * j - (k * (a[i] | a[j])), ans);
        }
      }
      for (long long i = n; i > max(0LL, n - 220LL); i--) {
        for (long long j = i - 1; j > max(0LL, n - 220LL); j--) {
          ans = max(i * j - (k * (a[i] | a[j])), ans);
        }
      }
      ans = max(ans, n * (n - 1) - (k * (a[n - 1] | a[n])));
      cout << ans << '\n';
    }
    return 0;
  }
  