#include <bits/stdc++.h>
using namespace std;
int t;
long long a[100003];
long long mn;
long long n, k;
void pr() { printf("%lld\n", mn); }
int main() {
  scanf("%d", &t);
  while (t--) {
    mn = -1e15;
    int gheirogheri__f3h804igh3g__fh93h4tgf4ig654465f265f__f423hui934hygf943515 =
        4565464;
    scanf("%lld%lld", &n, &k);
    double
        ogerhgheorgherhg__gehrigh30h4g0f34__fih32hf4if5454d324d2432d424f2424f44234 =
            2345.5445654;
    for (int i = 0; i < k; ++i)
      ++gheirogheri__f3h804igh3g__fh93h4tgf4ig654465f265f__f423hui934hygf943515;
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    long long
        gfjerop54504tih83g0934g340hg934hg934g435rg4rge4sg24seg24reag2442gfawgegr32a44eag =
            342535;
    for (long long i = n; i > 0; --i) {
      for (long long j = i - 1; j > 0; --j) {
        if (i * j <= mn) break;
        mn = max(mn, 1ll * i * j - k * (a[i] | a[j]));
      }
    }
    int giehgrhg34905g543__f3i94gh34i0g9oh__f3g49ig94354fg5344g544357g4g34g6583g4657fg63w4e7t743t64768 =
        32454;
    pr();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int t;
long long a[100003];
long long mn;
long long n, k;
void pr() { printf("%lld\n", mn); }
int main() {
  scanf("%d", &t);
  while (t--) {
    mn = -1e15;
    int gheirogheri__f3h804igh3g__fh93h4tgf4ig654465f265f__f423hui934hygf943515 =
        4565464;
    scanf("%lld%lld", &n, &k);
    double
        ogerhgheorgherhg__gehrigh30h4g0f34__fih32hf4if5454d324d2432d424f2424f44234 =
            2345.5445654;
    for (int i = 0; i < k; ++i)
      ++gheirogheri__f3h804igh3g__fh93h4tgf4ig654465f265f__f423hui934hygf943515;
    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    long long
        gfjerop54504tih83g0934g340hg934hg934g435rg4rge4sg24seg24reag2442gfawgegr32a44eag =
            342535;
    for (long long i = n; i > 0; --i) {
      for (long long j = i - 1; j > 0; --j) {
        if (i * j <= mn) break;
        mn = max(mn, 1ll * i * j - k * (a[i] | a[j]));
      }
    }
    int giehgrhg34905g543__f3i94gh34i0g9oh__f3g49ig94354fg5344g544357g4g34g6583g4657fg63w4e7t743t64768 =
        32454;
    pr();
  }
  return 0;
}

```