description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  clock_t startTime;
  double getCurrentTime() {
    return (double)(clock() - startTime) / CLOCKS_PER_SEC;
  }
  template <typename T, typename U>
  inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
    _out << _p.first << " " << _p.second;
    return _out;
  }
  template <typename T, typename U>
  inline istream &operator>>(istream &_in, pair<T, U> &_p) {
    _in >> _p.first >> _p.second;
    return _in;
  }
  template <typename T>
  inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
    if (_v.empty()) return _out;
    _out << _v.front();
    for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) _out << ' ' << *_it;
    return _out;
  }
  template <typename T>
  inline istream &operator>>(istream &_in, vector<T> &_v) {
    for (auto &_i : _v) _in >> _i;
    return _in;
  }
  template <typename T>
  inline ostream &operator<<(ostream &_out, const set<T> &_s) {
    if (_s.empty()) return _out;
    _out << *_s.begin();
    for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
    return _out;
  }
  template <typename T>
  inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
    if (_s.empty()) return _out;
    _out << *_s.begin();
    for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
    return _out;
  }
  template <typename T>
  inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
    if (_s.empty()) return _out;
    _out << *_s.begin();
    for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
    return _out;
  }
  template <typename T>
  inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
    if (_s.empty()) return _out;
    _out << *_s.begin();
    for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
    return _out;
  }
  template <typename T, typename U>
  inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
    if (_m.empty()) return _out;
    _out << "{\"" << _m.begin()->first << "\", \"" << _m.begin()->second << "\"}";
    for (auto _it = ++_m.begin(); _it != _m.end(); ++_it)
      _out << ", { \"" << _it->first << "\", \"" << _it->second << "\"}";
    return _out;
  }
  template <typename T, typename U>
  inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
    if (_m.empty()) return _out;
    _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
    for (auto _it = ++_m.begin(); _it != _m.end(); ++_it)
      _out << ", (" << _it->first << ": " << _it->second << ')';
    return _out;
  }
  template <typename T>
  void out(const vector<T> &a, int n) {
    for (int i = 1; i <= n; ++i) {
      cout << a[i] << ' ';
    }
    cout << "\n";
  }
  template <typename T>
  void out(const vector<vector<T>> &a, int n, int m) {
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        cout << a[i][j] << ' ';
      }
      cout << '\n';
    }
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n + 1);
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    long long ans = -1e15;
    for (long long i = 1; i <= n; ++i) {
      for (long long j = i - 1; j >= max(1LL, i - k); --j) {
        ans = max(ans, i * j - k * (a[i] | a[j]));
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    startTime = clock();
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  