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.*;
  public class cobb {
      public static void main(String[] args) {
          Scanner sc =new Scanner(System.in);
          int query= sc.nextInt();
          for(int a=0;a<query;a++){
              int n=sc.nextInt();
              int k=sc.nextInt();
              int[] array=new int[n];
              for(int b=0;b<n;b++){
                  array[b]= sc.nextInt();
              }
              int p;
              long value;
              long max =2-k*(array[0] | array[1]);
              int l;
              if(n-2*k-1>0){
                  l=n-2*k-1;
              }else{
                  l=0;
              }
                  for(int i=l;i<n;i++){
                      for(int j=i+1;j<n;j++){                   
                          p=(array[i] | array[j]);
                          value=((long)(i+1)*(j+1)) - (long)k*p; 
                          if (value>max){
                              max=value;
                          }                 
                      } 
                     
                  }
              
              
              System.out.println(max);
          }
          
      }
  }
  