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.io.*;
import java.util.*;

public class B {

    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        try{
            int t = Integer.parseInt(br.readLine());
            while(t-->0){
                String s1[] = br.readLine().split(" ");
                int n = Integer.parseInt(s1[0]);
                int k = Integer.parseInt(s1[1]);
                int a[] = new int[n];
                String s2[] = br.readLine().split(" ");
                for(int i=0; i<n; i++){
                    a[i] = Integer.parseInt(s2[i]);
                }
                long ans = Long.MIN_VALUE;
                for(int i=n-1; i>=Math.max(0,n-500); i--){
                    for(int j=i-1; j>=Math.max(0, n-500); j--){
                        long temp = (long)((i+1)*(j+1)) - (long)(k*(a[i]|a[j]));
                        ans = Math.max(ans,temp);
                    }
                }
                bw.write(ans+"\n");
            }
            bw.flush();
            bw.close();
        }catch(Exception e){
            return;
        }
    }

}