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.*;

import static java.lang.Math.*;

public class B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
       int t = in.nextInt();
        for (int tt = 0; tt < t; tt++) {
            int n = in.nextInt();
            long k = in.nextLong();
            long[] arr = new long[n];
            Arrays.setAll(arr, i->in.nextLong());
            pw.println(solve(arr, n, k));
        }
        pw.close();
    }
    static long solve(long[] arr, int n, long k){
        long till = 100;
        int began = 0;
        if (n > till) {
            began = n - 100;
        }
        long res = Long.MIN_VALUE;
        for (long i = began; i < n; i++) {
            for (long j = i + 1; j < n; j++) {
                long val = (i + 1) * (j + 1) - k * (arr[(int)i] | arr[(int)j]);
              //  debug(val);
                res = max(val, res);
            }
        }
        return res;
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }

}
