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 {
    static Scanner sc=new Scanner(System.in);
    public static void main(String[] args) {
        int t=sc.nextInt();
        while (t-->0) {
          int n=sc.nextInt();
          int k=sc.nextInt();
          int[] a=new int[n];
          int ind=-1,ct=100;
          for (int i=0;i<n;i++ ) {
            a[i]=sc.nextInt();

          }
          for (int i=0;i<n ;i++ ) {
            if(a[i]==0){
              ind=i;
              break;
            }
            if(getLeast(a[i])<ct){
              ct=getLeast(a[i]);
              ind=i;
            }
          }
          long ans=Long.MIN_VALUE;
          // for (int i=0;i<n-1;i++) {
          //   ans=Math.max((long)(i+1)*n-(long)k*(a[n-1]|a[i]),ans);
          // }
          // for (int i=0;i<n;i++) {
          //   if(i==ind)
          //     continue;
          //   ans=Math.max((long)(i+1)*(ind+1)-(long)k*(a[ind]|a[i]),ans);
          // }
          for(int i=1;i<n;i++)
            {
                ans=Math.max(ans,(i*(i+1)-k*(a[i-1] | a[i])));
          }
          System.out.println(ans);
        }
      }  
      public static  int getLeast(int n){
        int ct=0;
        while(n>0 && (n&1)==0){
          ct++;
          n>>=1;
        }
        return ct;
      }
  }
