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: import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  public class B {
      public static void main(String[] args) {
          Scanner in = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
         int t = in.nextInt();
          for (int tt = 0; tt < t; tt++) {
              int n = in.nextInt();
              long k = in.nextLong();
              long[] arr = new long[n];
              Arrays.setAll(arr, i->in.nextLong());
              pw.println(solve(arr, n, k));
          }
          pw.close();
      }
      static long solve(long[] arr, int n, long k){
          long till = 500;
          int began = 0;
          if (n > till) {
              began = n - 500;
          }
          long res = Long.MIN_VALUE;
          for (long i = began; i < n; i++) {
              for (long j = i + 1; j < n; j++) {
                  long val = (i + 1) * (j + 1) - k * (arr[(int)i] | arr[(int)j]);
                //  debug(val);
                  res = max(val, res);
              }
          }
          return res;
      }
  
      static void debug(Object... obj) {
          System.err.println(Arrays.deepToString(obj));
      }
  
  }
  