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 B {
  	static long[] arr;
  	static int n;
  	static long k;
  
  	public static void main(String[] args) throws IOException {
  		Kattio io = new Kattio();
  
  		int t = io.nextInt();
  		for (int ii=0; ii<t; ii++) {
  			n = io.nextInt();
  			k = io.nextLong();
  
  			arr = new long[n];
  
  			for (int i=0; i<n; i++) {
  				long x = io.nextLong();
  				arr[i] = x;
  			}
  
  			long ans = Integer.MIN_VALUE;
  
  			if (n <= 1000) {
  				for (int i=0; i<n; i++) {
  					for (int j=i+1; j<n; j++) {
  						//System.out.println(f(i, j));
  						//System.out.println(f(i, j));
  						ans = Math.max(ans, f(i, j));
  					}
  				}
  				io.println(ans);
  				continue;
  			}
  
  			if (n>1000) {
  				for (int i=0; i<1000; i++) {
  					for (int j=i+1; j<1000; j++) {
  						long ret = f(n-1-i, n-1-j);
  						ans = Math.max(ans, ret);
  					}
  				}
  
  				io.println(ans);
  				continue;
  			}
  		}
  		io.close();
  	}
  
  	static long f(int a, int b) {
  
  		long ret = (long) (a+1)* (long) (b+1) - (k* (arr[a] | arr[b]));
  		return ret; 
  	}
  
  	static class Kattio extends PrintWriter {
  		private BufferedReader r;
  		private StringTokenizer st;
  
  		// standard input
  		public Kattio() { this(System.in, System.out); }
  		public Kattio(InputStream i, OutputStream o) {
  			super(o);
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  		// USACO-style file input
  		public Kattio(String problemName) throws IOException {
  			super(new FileWriter(problemName + ".out"));
  			r = new BufferedReader(new FileReader(problemName + ".in"));
  		}
  
  		// returns null if no more input
  		public String next() {
  			try {
  				while (st == null || !st.hasMoreTokens())
  					st = new StringTokenizer(r.readLine());
  				return st.nextToken();
  			} catch (Exception e) { }
  			return null;
  		}
  
  		public int nextInt() { return Integer.parseInt(next()); }
  		public double nextDouble() { return Double.parseDouble(next()); }
  		public long nextLong() { return Long.parseLong(next()); }
  	}
  }