#P1554B. Cobb

    ID: 1953 Type: RemoteJudge 1000ms 256MiB Tried: 0 Accepted: 0 Difficulty: 5 Uploaded By: Tags>bitmasksbrute forcegreedymath*1700

Cobb

Description

You are given nn integers a1,a2,,ana_1, a_2, \ldots, a_n and an integer kk. Find the maximum value of ijk(aiaj)i \cdot j - k \cdot (a_i | a_j) over all pairs (i,j)(i, j) of integers with 1i<jn1 \le i < j \le n. Here, | is the bitwise OR operator.

The first line contains a single integer tt (1t100001 \le t \le 10\,000)  — the number of test cases.

The first line of each test case contains two integers nn (2n1052 \le n \le 10^5) and kk (1kmin(n,100)1 \le k \le \min(n, 100)).

The second line of each test case contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n (0ain0 \le a_i \le n).

It is guaranteed that the sum of nn over all test cases doesn't exceed 31053 \cdot 10^5.

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

Input

The first line contains a single integer tt (1t100001 \le t \le 10\,000)  — the number of test cases.

The first line of each test case contains two integers nn (2n1052 \le n \le 10^5) and kk (1kmin(n,100)1 \le k \le \min(n, 100)).

The second line of each test case contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n (0ain0 \le a_i \le n).

It is guaranteed that the sum of nn over all test cases doesn't exceed 31053 \cdot 10^5.

Output

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

Sample Input 1

4
3 3
1 1 3
2 2
1 2
4 3
0 1 2 3
6 6
3 2 0 0 5 6

Sample Output 1

-1
-4
3
12

Note

Let f(i,j)=ijk(aiaj)f(i, j) = i \cdot j - k \cdot (a_i | a_j).

In the first test case,

  • f(1,2)=12k(a1a2)=23(11)=1f(1, 2) = 1 \cdot 2 - k \cdot (a_1 | a_2) = 2 - 3 \cdot (1 | 1) = -1.
  • f(1,3)=13k(a1a3)=33(13)=6f(1, 3) = 1 \cdot 3 - k \cdot (a_1 | a_3) = 3 - 3 \cdot (1 | 3) = -6.
  • f(2,3)=23k(a2a3)=63(13)=3f(2, 3) = 2 \cdot 3 - k \cdot (a_2 | a_3) = 6 - 3 \cdot (1 | 3) = -3.

So the maximum is f(1,2)=1f(1, 2) = -1.

In the fourth test case, the maximum is f(3,4)=12f(3, 4) = 12.