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.*;
public class Solution
{
	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in); 
		int test= sc.nextInt();
		for(int xix=1;xix<=test;xix++)
		{
			int n=sc.nextInt(); int k=sc.nextInt();
			int[]a=new int[n];
			long[]r=new long[n];
			for(int x=0;x<n;x++)
			{
				a[x]=sc.nextInt();
				r[x]=x+1;
			}
			long max=r[1]*r[0]-k*(a[1]|a[0]);
			for(int i=n-1;i>=2;i--)
			{
				if(r[i]*r[i]-r[i]<=max)
				{
					break;
				}
				for(int j=i-1;j>=0;j--)
				{
					if(r[i]*r[j]<=max)
					{
						break;
					}
					if(r[i]*r[j]-k*(a[i]|a[j])>max)
					{
						max=r[i]*r[j]-k*(a[i]|a[j]);
					}
				}
			}
			System.out.println(max);
		}
		sc.close();
	}
}