#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
long long a[N];
signed main() {
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    long long ans = -4e18;
    for (int i = 1; i <= n; ++i) {
      for (int j = i - 1; j >= max(i - 200, 1); --j) {
        ans = max(ans, (long long)i * j - (long long)k * (a[i] | a[j]));
      }
    }
    printf("%lld\n", ans);
  }
  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 int N = 1e5 + 7;
long long a[N];
signed main() {
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    long long ans = -4e18;
    for (int i = 1; i <= n; ++i) {
      for (int j = i - 1; j >= max(i - 200, 1); --j) {
        ans = max(ans, (long long)i * j - (long long)k * (a[i] | a[j]));
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```