Description: You are given $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ and an integer $$$k$$$. Find the maximum value of $$$i \cdot j - k \cdot (a_i | a_j)$$$ over all pairs $$$(i, j)$$$ of integers with $$$1 \le i < j \le n$$$. Here, $$$|$$$ is the bitwise OR operator. Input Format: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ ($$$2 \le n \le 10^5$$$) and $$$k$$$ ($$$1 \le k \le \min(n, 100)$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le n$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$. Output Format: For each test case, print a single integer — the maximum possible value of $$$i \cdot j - k \cdot (a_i | a_j)$$$. Note: Let $$$f(i, j) = i \cdot j - k \cdot (a_i | a_j)$$$. In the first test case, - $$$f(1, 2) = 1 \cdot 2 - k \cdot (a_1 | a_2) = 2 - 3 \cdot (1 | 1) = -1$$$. - $$$f(1, 3) = 1 \cdot 3 - k \cdot (a_1 | a_3) = 3 - 3 \cdot (1 | 3) = -6$$$. - $$$f(2, 3) = 2 \cdot 3 - k \cdot (a_2 | a_3) = 6 - 3 \cdot (1 | 3) = -3$$$. So the maximum is $$$f(1, 2) = -1$$$. In the fourth test case, the maximum is $$$f(3, 4) = 12$$$.