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.
/*==========================================================================
*   	AUTHOR:    RonWonWon
*   	CREATED:   30.07.2021 02:10:12        
/*==========================================================================*/
import java.io.*;
import java.util.*;

public class B {
	
	public static void main(String[] args) throws IOException {
		/*
    in.ini();
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
    */
    long startTime = System.nanoTime();
    int t = in.nextInt(), T = 1;  
    while(t-->0) {
      int n = in.nextInt(); long k = in.nextInt();
      int a[] = in.readArray(n);
      int l = n - (int)(n - Math.sqrt((long)n*n - k*n));
      long mx = Long.MIN_VALUE;
      for(int i=l;i<n;i++){
      	for(int j=i+1;j<n;j++){
      		mx = Math.max(mx, (long)(i+1)*(j+1) - k*(a[i]|a[j]));
      	}
      }
      out.println(mx);
   		//out.println("Case #"+T+": "+ans); T++;
    }
    /*
    startTime = System.nanoTime() - startTime;
    System.err.println("Time: "+(startTime/1000000));
    */
    out.flush();
	}
	
	static FastScanner in = new FastScanner();
  static PrintWriter out = new PrintWriter(System.out);
	static int oo = Integer.MAX_VALUE;
	static long ooo = Long.MAX_VALUE;
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		void ini() throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		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());
		}
		
		long[] readArrayL(int n) {
			long a[] = new long[n];
			for(int i=0;i<n;i++) a[i] = nextLong();
			return a;
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
		
		double[] readArrayD(int n) {
			double a[] = new double[n];
			for(int i=0;i<n;i++) a[i] = nextDouble();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
	
	static void ruffleSortL(long[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n);
			long temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
}
