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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;



import static javax.swing.UIManager.get;
import static javax.swing.UIManager.getString;


public class Main {


    static class Pair implements Comparable<Pair>{
        int x=0;
        int y=0;
        public Pair(int x1,int y1)
        { x=x1;
            y=y1;
        }


        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    static boolean checkPallindrome(int n)
    {    ArrayList<Integer> list = new ArrayList<>();
        while(n>0)
        { list.add(n%10);
            n/=10;
        }
        int low=0,high=list.size()-1;
        while(low<=high)
        {     if(list.get(low)!=list.get(high))
            return false;
            low++;
            high--;
        }
        return true;
    }


    static boolean check(String s)
    {

        int low=0,high=s.length()-1;
        while(low<=high)
        {     if(s.charAt(low)!=s.charAt(high))
            return false;
            low++;
            high--;
        }
        return true;
    }

    class Node implements Comparable<Node>{
      int x=0,y=0,z=0;
        @Override
        public int compareTo(Node o) {
            return this.y-o.y;
        }
    }

    static  int min=Integer.MAX_VALUE;
    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
//(1)very very important**(never take the first problem for granted, always check the test cases) take 5 minutes more and check the edge cases
        // 5 minutes will not decreases rating as much as a wrong submission does it is easy u just think with an open mind and u will surely get the answer
        //(2)let ur brain consume the problem don't just jump to the solution. after reading the problem take a pause 1 minute
        //(3)go through the example test cases and also at least two of ur own test cases.Think of testcases which are difficult(edge cases).dry run ur concept

        int t = sc.nextInt();
        while (t-- > 0) {
          int n =sc.nextInt();
            int k =sc.nextInt();
          int[] a =sc.readArray(n);

    long max=Integer.MIN_VALUE;
    if(n-2*k-1>=0) {
        for (int i = n - 2 * k; i <= n; i++)
            for (int j = i + 1; j <= n; j++) {
                int temp = a[i - 1]|a[j - 1];
                max = Math.max(max, (i * j - k * temp));
            }
    }
    else
    { for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++) {
            int temp = a[i - 1]|a[j - 1];
            max = Math.max(max, (i * j - k * temp));

        }

    }
     System.out.println(max);
        }



    }



// syntax of conditional operator y=(x==1)?1:0;
//Things to check when u r getting wrong answer
// 1-  check the flow of the code
//2- If ur stuck read the problem once again
//3- before submitting always check the output format of ur code
//4- don't check standings until problem B is done
//5- if u r thinking ur concept is correct but still u r getting wrong answer try to implement it in another way
//6- By default, java interpret all numeral literals as 32-bit integer values.
// If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.    example   long a = 600851475143L

//All the functions



    static void debug(String s)
    { System.out.println(s);

    }

    //collections.sort use merge sort instead of quick sort but arrays.sort use quicksort whose worst time complexity is O(n^2)
    static int[] sort(int[] a)
    {         ArrayList<Integer> list = new ArrayList<>();
        for(int i=0;i<a.length;i++)
            list.add(a[i]);

        Collections.sort(list);
        int ind=0;
        for(int x:list)
            a[ind++]=x;
        return a;

    }

    //function to print an array for debugging
    static void print(int[] a) {
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }

    static void printc(char[] a) {
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }


    //normal gcd function, always put the greater number as a and the smaller number as b
    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);

    }

    static int lcm(int a,int b)
    {
        return (a*b)/gcd(a,b);
    }


    //to find gcd and lcm for numbers of long data type
    static long gcdl(long a, long b) {
        if (b == 0)
            return a;
        return gcdl(b, a % b);

    }

    static long lcml(long a,long b)
    {
        return (a*b)/gcdl(a,b);
    }


    //Input Reader to read faster input
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

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

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


        int[] readArray(int n)
        {  int[] a = new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();

            return a;
        }
    }



}

































































