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) throws Exception {
        PrintWriter pw = new PrintWriter(System.out);
        int test = nextInt();
        for (int t = 1; t <= test; t++) {
            int n = nextInt(), k = nextInt();
            int[] arr = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                arr[i] = nextInt();
            }
            pw.println(solve(n, k, arr));
        }

        pw.close();
    }

    static long solve(int n, long k, int[] arr) {
        int N = getLog(n);
        Pair[] pairs = new Pair[1 << N];
        Pair[] dp = new Pair[1 << N];

        for (int i = 0; i < pairs.length; i++) {
            pairs[i] = new Pair();
            dp[i] = new Pair();
        }

        for (int i = 1; i <= n; i++) {
            pairs[arr[i]].add(i);
        }

        for (int i = 0; i < (1 << N); ++i) {
            dp[i].add(pairs[i]);
        }
        for (int i = 0; i < N; ++i) {
            for (int mask = 0; mask < (1 << N); ++mask) {
                if ((mask & (1 << i)) != 0) {
                    dp[mask].add(dp[mask ^ (1 << i)]);
                }
            }
        }

        long res = Long.MIN_VALUE;
        for (int i = 0; i < dp.length; i++) {
            if (dp[i].size() < 2) {
                continue;
            }
            long x = (long) dp[i].first() * dp[i].last();
            long y = k * i;
            res = max(res, x - y);
        }

        return res;
    }

    static class Pair {
        int[] arr = {0, 0};

        void add(int x) {
            if (arr[1] < x) {
                arr[0] = arr[1];
                arr[1] = x;
            } else if (arr[0] < x) {
                arr[0] = x;
            }
        }

        int size() {
            if (arr[1] != 0 && arr[0] != 0) {
                return 2;
            } else {
                return 0;
            }
        }

        int first() {
            return arr[0];
        }

        int last() {
            return arr[1];
        }

        void add(Pair p) {
            for (int x : p.arr) {
                add(x);
            }
        }
    }

    static void trim(TreeSet<Integer> set) {
        while (set.size() > 2) {
            set.pollFirst();
        }
    }

    static int getLog(int n) {
        int i = 0;
        while ((1 << i) <= n) {
            i++;
        }
        return i;
    }

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

    static int nextInt() {
        try {
            boolean in = false;
            int res = 0;
            for (; ; ) {
                int b = System.in.read() - '0';
                if (b >= 0) {
                    in = true;
                    res = 10 * res + b;
                } else if (in)
                    return res;
            }
        } catch (IOException e) {
            throw new Error(e);
        }
    }
}