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 javax.swing.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class A {
    static ArrayList<Integer> [] adj;
    static ArrayList<Integer> temp;
    static int mod = (int) 1e9+7;
    static boolean[] vis = new boolean[(int)1e5];

    public static void main(String[] args) throws IOException {
        FastScanner s=new FastScanner();
        int t = s.nextInt();
//        int t = 1;
        while(t-->0){
            int n = s.nextInt();
            int k = s.nextInt();
            int[] a = s.readArray(n);
            long ans=Integer.MIN_VALUE;
            for(int i=n-1;i>=n-k-1 && i>=0;i--){
                for(int j=i-1;j>=n-k-1 && j>=0;j--){
                    ans = Math.max(ans,(i+1)*(long)(j+1)-k*(long)(a[i] | a[j]));
                }
            }
            System.out.println(ans);

        }
    }
//Important
//Dont't stick on same approach
//Check long




    static void dfs(int i){
        vis[i]=true;
        temp.add(i);
        List<Integer> Y = adj[i];
        for(int x:Y){
            if(!vis[x])dfs(x);
        }
    }

    static long gcd(long a, long b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void sortr(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l, Collections.reverseOrder());
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}