Codeforces Round 735 (Div. 2) |
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Finished |

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bitmasks

brute force

greedy

math

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B. Cobb

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputYou 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

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

For each test case, print a single integer — the maximum possible value of $$$i \cdot j - k \cdot (a_i | a_j)$$$.

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 $$$f(i, j) = i \cdot j - k \cdot (a_i | a_j)$$$.

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 $$$f(3, 4) = 12$$$.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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