#include <bits/stdc++.h>
using namespace std;
const long long oo = 1e18;
const long long N = 1e6 + 50;
const long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int test_cases = 1;
  cin >> test_cases;
  while (test_cases--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 10];
    long long ans = -oo;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i < min(n - 1, 220LL); i++) {
      for (long long j = i + 1; j <= min(n, 220LL); j++) {
        ans = max(i * j - (k * (a[i] | a[j])), ans);
      }
    }
    for (long long i = n; i > max(0LL, n - 220LL); i--) {
      for (long long j = i - 1; j > max(0LL, n - 220LL); j--) {
        ans = max(i * j - (k * (a[i] | a[j])), ans);
      }
    }
    ans = max(ans, n * (n - 1) - (k * (a[n - 1] | a[n])));
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long oo = 1e18;
const long long N = 1e6 + 50;
const long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int test_cases = 1;
  cin >> test_cases;
  while (test_cases--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 10];
    long long ans = -oo;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i < min(n - 1, 220LL); i++) {
      for (long long j = i + 1; j <= min(n, 220LL); j++) {
        ans = max(i * j - (k * (a[i] | a[j])), ans);
      }
    }
    for (long long i = n; i > max(0LL, n - 220LL); i--) {
      for (long long j = i - 1; j > max(0LL, n - 220LL); j--) {
        ans = max(i * j - (k * (a[i] | a[j])), ans);
      }
    }
    ans = max(ans, n * (n - 1) - (k * (a[n - 1] | a[n])));
    cout << ans << '\n';
  }
  return 0;
}

```