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

public class _G15 {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            long [] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = sc.nextInt();;
            if (n <= (int) 3e2) {
                // naive
                long ans = Long.MIN_VALUE;
                for (int i = 0; i < n; i++) {
                    for (int j = i + 1; j < n; j++) {
                        ans = Math.max(ans, ((long) i + 1) * ((long)j + 1) - k * (a[i] | a[j]));
                    }
                }
                out.println(ans);
            } else {
                int MAX = (int) 2e2;
                long ans = Long.MIN_VALUE;
                for (int i = 0; i < MAX; i++) {
                    for (int j = i + 1; j < MAX; j++) {
                        ans = Math.max(ans, ((long) n - i) * ((long) n - j) - k * (a[n - 1 - i] * a[n - 1 - j]));
                    }
                }
            }
        }
        out.close();
    }


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

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


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }


    }

}