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.
import java.util.*;
import java.io.*;
public class Solution{
    public static void main(String arg[]){
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t > 0)
        {
		    int i, j, n = scan.nextInt();
			long p, q, r = Long.MIN_VALUE, k = scan.nextLong();
			long [] nums = new long[n];
			for (i = 0; i < n; ++i)
			{
			nums[i] = scan.nextLong();
			}
			p = 1L;
			q = 1L;
            for (i = 0; i < n; ++i)
			{
			q = n - 2*k;
			for (j = Math.max(i + 1, (int)(n - 2*k - 1)); j < n; ++j)
			{
			r = Math.max(r, p*q - k*(nums[i]|nums[j]));
			q++;
			}
			p++;
			}
 			System.out.println(r);
            t--;
        }
    }
}