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: # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// PYTHON IS THE BEST ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  import sys,os,io
  from sys import stdin
  import math 
  from collections import defaultdict
  from heapq import heappush, heappop, heapify
  from bisect import bisect_left , bisect_right
  from io import BytesIO, IOBase
  BUFSIZE = 8192
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  
  
  alphabets = list('abcdefghijklmnopqrstuvwxyz')
  
  
  
  #for deep recursion__________________________________________-
  from types import GeneratorType
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to = stack[-1].send(to)
              return to
  
      return wrappedfunc
  
  def ncr(n, r, p):  
      num = den = 1
      for i in range(r):
          num = (num * (n - i)) % p
          den = (den * (i + 1)) % p
      return (num * pow(den,p - 2, p)) % p
  
  def primeFactors(n): 
      l = []
      while n % 2 == 0: 
          l.append(2)
          n = n / 2
      for i in range(3,int(math.sqrt(n))+1,2): 
          while n % i== 0: 
              l.append(int(i))
              n = n / i 
      if n > 2: 
          l.append(n)
      # c = dict(Counter(l))
      return list(set(l))
      # return c
  
  def power(x, y, p) : 
  	res = 1
  	x = x % p 
  	if (x == 0) : 
  		return 0
  	while (y > 0) : 
  		if ((y & 1) == 1) : 
  			res = (res * x) % p 
  		y = y >> 1	 # y = y/2 
  		x = (x * x) % p 		
  	return res 
  
  #____________________GetPrimeFactors in log(n)________________________________________
  def sieveForSmallestPrimeFactor():
      MAXN = 100001
      spf = [0 for i in range(MAXN)]
      spf[1] = 1
      for i in range(2, MAXN):
          spf[i] = i
      for i in range(4, MAXN, 2):
          spf[i] = 2
      for i in range(3, math.ceil(math.sqrt(MAXN))):
          if (spf[i] == i):
              for j in range(i * i, MAXN, i): 
                  if (spf[j] == j):
                      spf[j] = i
      return spf
  def getPrimeFactorizationLOGN(x):
      spf = sieveForSmallestPrimeFactor()
      ret = list()
      while (x != 1):
          ret.append(spf[x])
          x = x // spf[x]  
      return ret
  #____________________________________________________________
  
  
  
  def SieveOfEratosthenes(n): 
      #time complexity = nlog(log(n))
      prime = [True for i in range(n+1)]
      p = 2
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n+1, p):
                  prime[i] = False
          p += 1
      return prime
  def si():
      return input()
  def divideCeil(n,x):
      if (n%x==0):
          return n//x
      return n//x+1
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  
  
  # ///////////////////////////////////////////////////////////////////////////
  # //////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
  # ///////////////////////////////////////////////////////////////////////////
  if(os.path.exists('input.txt')):
      sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
      input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  
  def solve():
      n,k = li()
      a = li()
      ans = float('-inf')
      for i in range(n-1,max(-1,n-201),-1):
          for j in range(i-1,max(-1,n-201),-1):
              # print(i,j)
              x = (i+1)*(j+1) - (a[i]|a[j])*k 
              ans = max(ans,x)
      print(ans)
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  