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 java.io.*;
  
  public class B_1554 {
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		
  		int t = sc.nextInt();
  		while(t-->0) {
  			int n = sc.nextInt(), k = sc.nextInt();
  			int[] array = sc.nextIntArray(n);
  			
  			long max = Long.MIN_VALUE;
  			for(int i = n - 1; i >= Math.max(n - 1000, 0); i--)
  				for(int j = i - 1; j >= Math.max(n - 1000, 0); j--)
  					max = Math.max(max, 1l * (i + 1) * (j + 1) - k * (array[i]|array[j]));
  			
  			pw.println(max);
  		}
  		
  		pw.flush();
  	}
  
  	public static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextIntArray(int n) throws IOException {
  			int[] array = new int[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextInt();
  			return array;
  		}
  
  		public Integer[] nextIntegerArray(int n) throws IOException {
  			Integer[] array = new Integer[n];
  			for (int i = 0; i < n; i++)
  				array[i] = new Integer(nextInt());
  			return array;
  		}
  
  		public long[] nextLongArray(int n) throws IOException {
  			long[] array = new long[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextLong();
  			return array;
  		}
  
  		public double[] nextDoubleArray(int n) throws IOException {
  			double[] array = new double[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextDouble();
  			return array;
  		}
  
  		public static int[] shuffle(int[] a) {
  			int n = a.length;
  			Random rand = new Random();
  			for (int i = 0; i < n; i++) {
  				int tmpIdx = rand.nextInt(n);
  				int tmp = a[i];
  				a[i] = a[tmpIdx];
  				a[tmpIdx] = tmp;
  			}
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }
  