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.
#include <algorithm>   
#include <cstring>
#include <deque>　　     
#include <iostream>     
#include <list>　　　　 
#include <map>　　     
#include <queue>　　   
#include <set>　　　　 
#include <stack>　　    　　
#include <string>　　   
#include <vector>　　   

#define fi first
#define se second
#define cl(a, b) memset(a, b, sizeof(a))
#define rapido ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

using namespace std;

const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> PII;
typedef long long int ll;
const int N = 1e5 + 10;

ll a[N];
int n, k;

void solve()
{
	cin >> n >> k;
	for(int i = 1; i <= n; i ++ )
	cin >> a[i];
	
	ll res = - INF;
	if(n <= 1000)
	{
		for(int i = 1; i <= n; i ++ )
		{
			for(int j = i + 1; j <= n; j ++ )
			res = max(res, i * j - k * (a[i] | a[j]));
		}
	}
	else 
	{
		for(int i = n - 1000; i <= n; i ++ )
		{
			for(int j = i + 1; j <= n; j ++ )
			res = max(res, i * j - k * (a[i] | a[j]));
		}
	}
	cout << res << endl;
}

int main()
{
    rapido;
    int t;
    for(cin >> t; t; t -- )
    //for (t = 1; t; t -- )
        solve();
}