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.util.*;
      import static java.lang.Math.max;
      import static java.lang.Math.min;
      import static java.lang.Math.abs;
      import static java.lang.Math.ceil;
      import static java.lang.Math.max;
      import static java.lang.Math.pow;
      import static java.lang.System.out;
       
      public class Codechef {
      	static Scanner s = new Scanner(System.in);
       
      	public static void main(String[] args) {
      		int t = s.nextInt();
      		while(t-->0) {
      			int n = s.nextInt();
      			long k = s.nextLong();
      			long[] arr = new long[n];
      			for(int i = 0; i < n; i++) {
      				arr[i] = s.nextLong();
      			}
      			long ans = Integer.MIN_VALUE;
      			for(int i = n-1, c = 1; i >= 0 && c <= 300; i--,c++) {
      				for(int j = i+1; j < n; j++) {
      					long temp = ((i+1L)*(j+1L)) - (k*(arr[i]|arr[j]));
      					ans = max(temp, ans);
      				}
      			}			
      			System.out.println(ans);
      		}
      	}
       
      	public static int gcd(int a, int b)
      	{
      		if (b == 0)
      			return a;
      		return gcd(b, a % b);
      	}
       
       
      	public static void printarr(int[] arr) {
      		for(int i = 0; i < arr.length; i++) {
      			System.out.print(arr[i]+" ");
      		}
      	}
       
      	public static void printarr(int arr[][]) {
      		for(int i = 0 ; i < arr.length; i++) {
      			for(int j = 0; j < arr[0].length; j++) {
      				System.out.println(arr[i][j]+" ");
      			}
      			System.out.println();
      		}
      	}
       
      	public static void takeip(int arr[]) {
      		for(int i = 0; i < arr.length; i++) {
      			arr[i] = s.nextInt();
      		}
      	}
       
      	public static void takeip(int arr[][]) {
      		for(int i = 0 ; i < arr.length; i++) {
      			for(int j = 0; j < arr[0].length; j++) {
      				arr[i][j] = s.nextInt();
      			}
      		}
      	}
      }