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 Main {
  	public static void main(String[] args) {
  		InputStream inputStream = System.in;
  		OutputStream outputStream = System.out;
  		InputReader in = new InputReader(inputStream);
  		PrintWriter out = new PrintWriter(outputStream);
  		Task solver = new Task();
  		int tcCnt = in.nextInt();
  		for (int tc = 1; tc <= tcCnt; tc++)
  			solver.solve(tc, in, out);
  		out.close();
  	}
  
  	static class Task {
  		public void solve(int testNumber, InputReader in, PrintWriter out) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = in.nextInt();
  			}
  			long ans = Long.MIN_VALUE;
  			for (int i = n; i >= Math.max(1, n - 2*k); i--) {
  				for (int j = i-1; j >= Math.max(1, n - 2*k); j--) {
  					ans = Math.max(ans, 1l * i * j - 1l * k * (a[i-1] | a[j-1]));
  				}
  			}
  			out.println(ans);
  		}
  
  	}
  
  	static class InputReader {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch (IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  
  		public String nextToken() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				tokenizer = new StringTokenizer(nextLine());
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public int[] nextIntArray(int size) {
  			int[] arr = new int[size];
  			for (int i = 0; i < size; i++) {
  				arr[i] = nextInt();
  			}
  			return arr;
  		}
  
  	}
  }