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
  import math
  
  if sys.subversion[0] == "PyPy":
      import io, atexit
  
      sys.stdout = io.BytesIO()
      atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
  
      sys.stdin = io.BytesIO(sys.stdin.read())
      input = lambda: sys.stdin.readline().rstrip()
  
  t = int(raw_input())
  
  # python codesforces210710.py < input.txt
  for zz in range(t):
  
      str1 = raw_input()
      n, k = str1.split(' ')
      n = int(n)
      k = int(k)
      str2 = raw_input()
      lis = str2.split(' ')
      lis = map(int, lis)
  
      lvmi = max(n * (n-1.0) - 2 * k * n, 0)
      lvfori = int(max(lvmi ** 0.5 - 2, 0))
      mma = - k * n * n
  
      for i in range(lvfori, n):
          for j in range(i + 1, n):
              lp = k * (lis[i] | lis[j])
              canma = (i + 1) * (j + 1) - k * (lis[i] | lis[j])
              mma = max(canma, mma)
      print(mma)
  
  
  
  
  
  