#include <bits/stdc++.h>
using namespace std;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << '(' << p.f << ',' << ' ' << p.s << ')';
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << '[';
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ',';
    cout << v[i];
  }
  return cout << ']' << endl;
}
const long long mod = 1000000007;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > arr(n);
  long long ans = -1e15;
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i + 1;
  }
  reverse(arr.begin(), arr.end());
  for (int i = 0; i < min(n, 1000LL); i++) {
    for (int j = i + 1; j < min(n, 1000LL); j++) {
      ans = max(ans, arr[i].second * arr[j].second -
                         k * (arr[i].first | arr[j].first));
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << '(' << p.f << ',' << ' ' << p.s << ')';
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << '[';
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ',';
    cout << v[i];
  }
  return cout << ']' << endl;
}
const long long mod = 1000000007;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > arr(n);
  long long ans = -1e15;
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i + 1;
  }
  reverse(arr.begin(), arr.end());
  for (int i = 0; i < min(n, 1000LL); i++) {
    for (int j = i + 1; j < min(n, 1000LL); j++) {
      ans = max(ans, arr[i].second * arr[j].second -
                         k * (arr[i].first | arr[j].first));
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```