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[] ct=new int[n+5];
          Map<Integer,ArrayList<Integer>> map=new HashMap<>();
          // int ind=-1,ct=100;
          int mx=0,mi=n+5;
          for (int i=0;i<n;i++ ) {
            a[i]=sc.nextInt();
            ct[a[i]]++;
            mx=Math.max(mx,a[i]);
            mi=Math.min(mi,a[i]);
            if(map.containsKey(a[i])){
              ArrayList<Integer> temp=map.get(a[i]);
              temp.add(i);
              map.put(a[i],temp);
            }else{
              ArrayList<Integer> temp=new ArrayList<>();
              temp.add(i);
              map.put(a[i],temp);
            }
          }
          // 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;
          //   }
          // }
          int z=0;
          int[] x=new int[n];
          for (int i=0;i<ct.length;i++ ) {
            while (ct[i]>0) {
              x[z++]=i;
              ct[i]--;
            }
          }
          long ans=Long.MIN_VALUE;
          for(int i=n-1;i>=1;i--){
            for(int j=i-1;j>=0;j--){
              if(i*j<ans){
                break;
              }
              ans=Math.max((long)(i+1)*(j+1)-(long)k*(a[i]|a[j]),ans);
            }
          }
          System.out.println(ans);
        }
      }  
      public static  int getLeast(int n){
        int ct=0;
        while(n>0 && (n&1)==0){
          ct++;
          n>>=1;
        }
        return ct;
      }
  }
