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.math.BigInteger;
import java.util.*;

public class Agfg2 {
    public static void main(String[] args) {
        handmadeInput.FastReader in = new handmadeInput.FastReader();
        int t = in.nextInt();
        StringBuilder s = new StringBuilder("");
        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] qus = new int[n + 10];
            for (int i = 1; i <= n; i++) {
                qus[i] = in.nextInt();
            }
            long ans = Long.MIN_VALUE;
            int till = Math.max(1, n - 201);
            for (int i = n; i >= till; i--) {
                for (int j = i - 1; j >= till; j--) {
                    int temp = i * j - k * (qus[i] | qus[j]);
                    ans = Math.max(ans, temp);
                }

            }
            s.append(ans).append(" ");


        }
        System.out.println(s);

    }
}





// Additional Classes go here



class handmadeInput {
    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;
        }
    }
}

class UtilityFunctions {
    //! returns gcd of given numbers
   public int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

 public int primeFactors(int n) {
        HashMap<Integer, Integer> hashMap = new HashMap<>(n);
        int c = 0;

        // Print the number of 2s that divide n
        while (n % 2 == 0) {
//            if (hashMap.containsKey(2)) {
//                hashMap.replace(2, hashMap.get(2) + 1);
//            } else {
//                hashMap.put(2, 1);
//            }
            c++;
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            // While i divides n, print i and divide n
            while (n % i == 0) {
                //System.out.print(i + " ");

//                if (hashMap.containsKey(i)) {
//                    hashMap.replace(i, hashMap.get(i) + 1);
//                } else {
//                    hashMap.put(i, 1);
//                }
                c++;
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2) {

//            if (hashMap.containsKey(n)) {
//                hashMap.replace(n, hashMap.get(n) + 1);
//            } else {
//                hashMap.put(n, 1);
//            }
            c++;
        }

        //System.out.print(n);
        return c;
    }
}
