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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  //import java.math.BigInteger;
  import java.util.*;
  
  public class Agfg2 {
      public static void main(String[] args) {
          handmadeInput.FastReader in = new handmadeInput.FastReader();
          int t = in.nextInt();
          StringBuilder s = new StringBuilder("");
          while (t-- > 0) {
              int n = in.nextInt();
              int k = in.nextInt();
              int[] qus = new int[n + 10];
              for (int i = 1; i <= n; i++) {
                  qus[i] = in.nextInt();
              }
              long ans = Long.MIN_VALUE;
              int till = Math.max(1, n - 201);
              for (int i = n; i >= till; i--) {
                  for (int j = i - 1; j >= till; j--) {
                      long temp = (long) i * j - (long) k * (qus[i] | qus[j]);
                      ans = Math.max(ans, temp);
                  }
  
              }
              s.append(ans).append(" ");
  
  
          }
          System.out.println(s);
  
      }
  }
  
  
  
  
  
  // Additional Classes go here
  
  
  
  class handmadeInput {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  
  class UtilityFunctions {
      //! returns gcd of given numbers
     public int gcd(int a, int b)
      {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
  
   public int primeFactors(int n) {
          HashMap<Integer, Integer> hashMap = new HashMap<>(n);
          int c = 0;
  
          // Print the number of 2s that divide n
          while (n % 2 == 0) {
  //            if (hashMap.containsKey(2)) {
  //                hashMap.replace(2, hashMap.get(2) + 1);
  //            } else {
  //                hashMap.put(2, 1);
  //            }
              c++;
              n /= 2;
          }
  
          // n must be odd at this point.  So we can
          // skip one element (Note i = i +2)
          for (int i = 3; i <= Math.sqrt(n); i += 2) {
              // While i divides n, print i and divide n
              while (n % i == 0) {
                  //System.out.print(i + " ");
  
  //                if (hashMap.containsKey(i)) {
  //                    hashMap.replace(i, hashMap.get(i) + 1);
  //                } else {
  //                    hashMap.put(i, 1);
  //                }
                  c++;
                  n /= i;
              }
          }
  
          // This condition is to handle the case whien
          // n is a prime number greater than 2
          if (n > 2) {
  
  //            if (hashMap.containsKey(n)) {
  //                hashMap.replace(n, hashMap.get(n) + 1);
  //            } else {
  //                hashMap.put(n, 1);
  //            }
              c++;
          }
  
          //System.out.print(n);
          return c;
      }
  }
  