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 sys
  testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
  interactive = False
  if testing:
      cmd = sys.stdout
      from time import time
      start_time = int(round(time() * 1000)) 
      readAll = open(sys.argv[1], 'r').read
      sys.stdout = open(sys.argv[2], 'w')
  else:
      readAll = sys.stdin.read
  
  # ############ ---- I/O Functions ---- ############
  
  class InputData:
      def __init__(self):
          self.lines = readAll().split('\n')
          self.n = len(self.lines)
          self.ii = -1
      def input(self):
          self.ii += 1
          assert self.ii < self.n
          return self.lines[self.ii]
  
  flush = sys.stdout.flush
  if interactive and not testing:
      input = sys.stdin.readline
  else:
      inputData = InputData()
      input = inputData.input
  
  def intin():
      return(int(input()))
  def intlin():
      return(list(map(int,input().split())))
  def chrin():
      return(list(input()))
  def strin():
      return input()
  def lout(l, sep="\n", toStr=True):
      print(sep.join(map(str, l) if toStr else l))
  def dout(*args, **kargs):
      if not testing: return
      if args: print(args[0] if len(args)==1 else args)
      if kargs: print([(k,v) for k,v in kargs.items()])
  def ask(q):
      sys.stdout.write(str(q)+'\n')
      flush()
      return intin()
      
  # ############ ---- I/O Functions ---- ############
  
  # from math import ceil
  # from collections import defaultdict as ddict, Counter
  # from heapq import *
  # from Queue import Queue
  
  def main():
      n,k = intlin()
      a = intlin()
      ans = -10**9
      # b = [0]*(n+1)
      # c = [0]*(n+1)
      # for i in xrange(n-1,-1,-1):
      #     if b[a[i]] == 0:
      #         b[a[i]] = i+1
      #     elif c[a[i]] == 0:
      #         c[a[i]] = i+1
      # for i in xrange(n):
      #     if b[i]*c[i] != 0:
      #         ans = max(ans,b[i]*c[i] - k*i)
      for i in xrange(max(0,n-1000),n):
          for j in xrange(max(0,n-1000), n):
              if i == j:
                  continue
              ans = max(ans,(i+1)*(j+1)- k*(a[i]|a[j]))
      return ans
  
          
      
  anss = []
  for _ in xrange(intin()):
      anss.append(main())
      # anss.append("YES" if main() else "NO")
  lout(anss)
  
  if testing:
      sys.stdout = cmd
      print(int(round(time() * 1000))  - start_time)